Climbing the Mountain

Real Analysis Exercises III

a) Calculate:

$\displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k) \quad \mathrm{and} \quad \sum_{k=p}^{m}(u_k - u_{k+1})$

$\sum_{k=p}^{m}(u_{k+1}-u_k)=u_{p+1}-u_{p}+u_{p+2}-u_{p+1}+\ldots +u_{m+1}-u_{m}$

As we can see the first term cancels out with the fourth, the third with the sixth, and so on and all we are left with is the second and second last terms:

$\sum_{k=p}^{m}(u_{k+1}-u_k) = u_{m+1}-u_p$

$\sum_{k=p}^{m}(u_k - u_{k+1})= - \sum_{k=p}^{m}(u_{k+1}-u_k)= - (u_{m+1}-u_p)= u_p-u_{m+1}$

b) Calculate $\lim \sum_{k=1}^n\dfrac{1}{k(k+1)}$ using the previous result.

$\lim \sum_{k=1}^n\dfrac{1}{k(k+1)}= \lim \sum_{k=1}^n \left( \frac{1}{k}-\frac{1}{k+1} \right)$

Defining ${u_k=1/k}$ the previous sum can be written as

$\lim \sum_{k=1}^n \left( u_k-u_{k+1} \right)=\lim (u_1 - u_{n+1})= \lim \left(1-\frac{1}{n+1}\right)=1$

This last result apparently has a funny story. Mengoli was the first one to calculate $\lim \sum_{k=1}^n\dfrac{1}{k(k+1)}=1$ . At the time this happened people did research in mathematics (I’m using this term rather abusively) in a somewhat different vein. They didn’t rush to print what they found like today. Many times people held out their results for years while tormenting their rivals about what they found. This is exactly what Mengoli did. In the times he was around the theory of series wasn’t much developed, thus this result, that we can calculate without being particularly brilliant in Mathematics, was something to take note of. So, he wrote some letters to people saying that $\lim \sum_{k=1}^n \left( u_k-u_{k+1} \right)=\lim (u_1 - u_{n+1})= \lim \left(1-\frac{1}{n+1}\right)=1$ , but not how he concluded that. The other mathematicians he sent the result too didn’t know about his methods and all they could do was to add numbers up explicitly and the only thing they could see was that even though they could sum more and more terms the result was approaching ${1}$ . Of course this didn’t prove nothing since summing up a billion terms isn’t the same as summing an infinite number of terms.

c) Calculate $\sum_{k=0}^{n-1}(2k+1)$

In this exercise what we are calculating is the sum of ${n}$ consecutive odd numbers. This result was already known to the ancient Greeks and the result wasn’t nothing short to astounding to them. But enough with the talk already:

$\sum_{k=0}^{n-1}(2k+1)=\sum_{k=0}^{n-1}\left[ (k+1)^2-k^2\right] = \sum_{k=0}^{n-1}(u_{k+1}-u_k)$ with ${u_k=k^2}$

Using the now familiar formula $\sum_{k=0}^{n-1}(2k+1) = (n-1+1)^2-0^2=n^2$

An astounding result indeed. Just look to $\sum_{k=0}^{n-1}(2k+1)=n^2$ , interpret the result and try not to be as surprised as the ancient Greeks were.

a) Using 1.a) and ${a^k=a^k\dfrac{a-1}{a-1}\quad (a \neq 1)}$ calculate $\sum_{k=0}^{n-1} a^k$

${\begin{array}{rcl} \displaystyle \displaystyle \sum_{k=0}^{n-1} a^k &=& \displaystyle\sum_{k=0}^{n-1} \left[ a^k\frac{a-1}{a-1}\right]= \\ \displaystyle \frac{1}{a-1}\sum_{k=0}^{n-1}\left( a^{k+1}-a^k\right) &=& \displaystyle\frac{1}{a-1}(a^n-1)= \\ &=& \displaystyle\frac{a^n-1}{a-1} \end{array}}$

b) Using a) establish the Bernoulli inequality ${a^n-1 \geq n(a-1)}$ if ${a>0}$ and ${n \in \mathbb{Z}^+}$

If ${a=1}$ it is ${1-1=n(1-1) \Rightarrow 0=0}$ which is trivially true.

If ${n=1}$ it is ${a-1=a-1}$ which is trivially true.

For ${n \geq 2 }$ and ${a>1}$ it is:

$\sum_{k=0}^{n-1}a^k=1+a+a^2+\ldots+a^{n-1}>1+1+\ldots+1=n$

Thus ${\dfrac{a^n-1}{a-1}>n \Rightarrow a^n-1>n(a-1)}$ since ${a>1}$

Finally if ${0<a<1}$ it is

$\sum_{k=0}^{n-1} a^k = 1+a+a^2+\ldots+a^{n-1}<n$

Thus ${\dfrac{a^n-1}{a-1} < n \Rightarrow a^n - 1 > n(a-1)}$ since ${a<1}$

c) Use b) to calculate ${\lim a^n}$ if ${a>1}$ and then conclude that ${\lim a^n=0}$ if ${|a|<1}$ .

By b) it is ${a^n>n(a-1)+1 \Rightarrow \lim a^n \geq \lim \left( n(a-1)+1 \right)= +\infty }$ . Hence ${\lim a^n = +\infty \quad (a>1)}$

For the second part of the exercise we will calculate instead ${\lim |a^n|}$ since that we know that ${ u_n \rightarrow 0 \Leftrightarrow |u_n| \rightarrow 0}$

Let us make a change of variable ${t=1/a}$ . Thus ${|a|=|1/t|}$ and

${\lim |a^n|= \lim |1/t|^n=\dfrac{1}{\lim |t|^n}=\dfrac{1}{+\infty}=0}$

3. Consider the sequences ${u_n=\left( 1+\dfrac{1}{n} \right)^n }$ and ${v_n=\left( 1+\dfrac{1}{n} \right)^{n+1}}$

a) Calculate ${\dfrac{v_n}{v_{n+1}}}$ and ${\dfrac{u_{n+1}}{u_n}}$ . Then use Bernoulli’s inequality to show that ${v_n}$ is strictly decreasing and that ${u_n}$ is strictly increasing.

${\begin{array}{rcl} \dfrac{v_n}{v_{n+1}} &=& \dfrac{\left( 1+1/n \right)^{n+1}}{\left(1+1/(n+1)\right)^{n+2}} \\ \dfrac{\left(\dfrac{n+1}{n}\right)^{n+1}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}} &=& \dfrac{n}{n+1}\dfrac{\left(\dfrac{n+1}{n}\right)^{n+2}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}} \\ \dfrac{n}{n+1}\left( \dfrac{(n+1)^2}{n(n+2)} \right)^{n+2} &=& \dfrac{n}{n+1}\left( \dfrac{n^2+2n+1}{n(n+2)} \right)^{n+2} \\ \dfrac{n}{n+1}\left( \dfrac{n(n+2)+1}{n(n+2)} \right)^{n+2} &=& \dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} \end{array}}$

After having calculated ${v_n/v_{n+1}}$ we can use Bernoulli’s inequality, with ${a=1+\dfrac{1}{n(n+2)}}$ , to conclude that ${v_n}$ is strictly decreasing.

${\dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} > \dfrac{n}{n+1}\left(1 + \dfrac{n+2}{n(n+2)} \right)}$

${ \dfrac{n}{n+1}(1+1/n)=\dfrac{n}{n+1}\dfrac{n+1}{n}=1 }$

Thus ${v_n}$ is strictly decreasing.

With a similar technique we can prove that

$\displaystyle u_{n+1}/u_n=\dfrac{n+1}{n}\left( 1- \dfrac{1}{(n+1)^2}\right)^{n+1}$

After that by using Bernoulli’s inequality like in the previous example one can show that ${u_{n+1}/u_n>1}$ and thus ${u_n}$ is strictly increasing.

c) Using a) and b) and ${\lim u_n = e}$ prove the following inequalities ${(1+1/n)^n<e<(1+n)^{n+1}}$

${\lim v_n= \lim(1+1/n)^n(1+1/n)=e\times 1=e}$ and we already know that ${v_n}$ is decreasing so it is ${v_n<(1+1/n)^{n+1}}$

On the other hand ${u_n}$ is increasing and ${\lim u_n=e}$ so ${(1+1/n)^n<e}$ . Hence ${(1+1/n)^n<e<(1+1/n)^{n+1}}$

d) Use c) to prove that

$\displaystyle \frac{1}{n+1}<\log (n+1)-\log n <\frac{1}{n}$

$\begin{array}{rcl} (1+1/n)^n &<& e \Rightarrow \\ \Rightarrow n \log \left( \dfrac{n+1}{n} \right) &<& 1 \Rightarrow \\ \Rightarrow \log(n+1) - \log n &<& \dfrac{1}{n} \end{array}$

And now for the second part of the inequality:

$\begin{array}{rcl} e &<& (1+1/n)^{n+1} \Rightarrow \\ \Rightarrow 1 &<& (n+1)\log \left(\dfrac{n+1}{n}\right) \Rightarrow \\ \Rightarrow \dfrac{1}{n+1} &<& \log (n+1) -\log n \end{array}$

In conclusion it is ${ \dfrac{1}{n+1}<\log (n+1)- \log n < \dfrac{1}{n} }$

a) Using 3d) show that

$\displaystyle 1+\log k < (k+1)\log (k+1)-k\log k < 1+ \log(k+1)$

From

${ \begin{array}{rcl} \dfrac{1}{k+1} &<& \log (k+1) - \log k \Rightarrow \\ \Rightarrow 1 &<& (k+1)\log(k+1) - (k+1)\log k \Rightarrow \\ \Rightarrow 1+ \log k &<& (k+1)\log(k+1)-k \log k \end{array}}$

With a similar reasoning we can also prove that ${(k+1)\log(k+1)-l\log k< 1+ \log(k+1)}$ . Thus it is ${1+\log k< (k+1)\log(k+1)-k\log k< 1+ \log(k+1)}$

b) Sum the previous inequalities between ${1 \leq k \leq n-1}$ .

$\sum_{k=1}^{n-1}(1+ \log k)<\sum_{k=1}^{n-1} ((k+1)\log(k+1)-k \log k)<$

$<\sum_{k=1}^{n-1}(1+\log(k+1))$

Now

$\sum_{k=1}^{n-1} (1+ \log k) = \sum_{k=1}^{n-1}1+\sum_{k=1}^{n-1}\log k= n-1 +\sum_{k=1}^{n-1}\log k$

with $\sum_{k=1}^{n-1}\log k = \log 1 + \log2 +\ldots+\log(n-1)=\log((n-1)!)$

It also is $\sum_{k=1}^{n-1}((k+1)\log(k+1) - k\log k)= m\log n -\log 1 =n\log n$ (This is a Mengoli sum)

And $\sum_{k=1}^{n-1}(1+\log(k+1))=n-1+\log n!$

Thus it is ${n-1+\log(n-1)! < n\log n < n-1 \log n!}$

c) Conclude the following inequalities ${ n \log n -n +1<\log n! < n \log n -n+1+\log n}$ and establish Stirling’s approximation

$\displaystyle \log n! = n\log n -n +r_n\,\, \mathrm{with}\,\, e<C_n<en$

${ \begin{array}{rcl} n-1 + \log (n-1)! &<& n\log n \Rightarrow \\ \Rightarrow \log (n-1)! &<& n\log n -n+1 \Rightarrow \\ \Rightarrow \log n! &<& n\log n -n +1+\log n \end{array} }$

On the other hand

${ \begin{array}{rcl} n\log n &<& n-1 + \log n! \Rightarrow \\ \Rightarrow n\log n -n +1 &<& \log n! \end{array} }$

Thus ${n\log n -n +1 < \log n! <n\log n -n +1 +1log n}$ and from this follows ${1<\log n! -n\log n+n<1+\log n}$

Defining ${r_n=\log n! -n\log n+n}$ it is ${\log n! n\log n-n+r_n}$ with ${1<r_n<1+\log n}$

Show that ${\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$ and that ${\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$

We know that

$\begin{array}{lcr} \dfrac{1}{n+1}<\log(n+1)-\log n < \dfrac{1}{n} \\ \dfrac{1}{n+1}<\log\left( \dfrac{n+1}{n}\right) < \dfrac{1}{n} \\ \dfrac{1}{n+1} <\log\left( 1+\dfrac{1}{n}\right) <\dfrac{1}{n} \\ \dfrac{1/(n+1)}{1/n} < \dfrac{\log (1+1/n)}{1/n}<1 \\ \leq lim \dfrac{n}{n+1} \leq \lim \dfrac{\log (1+1/n)}{1/n} \leq \lim 1 \\ 1 \leq \lim \dfrac{\log (1+1/n)}{1/n} \leq 1 \end{array}$

Thus ${\lim \dfrac{\log (1+1/n)}{1/n}=1}$ and this equivalent to saying that ${\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$

Let ${u_n = \dfrac{\log (1+1/n)}{1/n}}$ . In this case it is ${\dfrac{\log (1+1/n^2)}{1/n^2}=u_{n^2}}$ . Since ${u_{n^2}}$ is a subsequence of ${u_n}$ we know that ${\lim u_{n^2}= \lim u_n}$ and so it also is ${\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$

6. Show that ${u_n \sim v_n}$ and ${v_n \sim w_n \Rightarrow u_n \sim w_n }$

By hypothesis it is ${u_n=h_n v_n}$ , ${v_n=t_n w_n}$ with ${h_n,t_n \rightarrow 1}$ . Substituting the second equality in the first we obtain ${u_n = h_n t_n w_n}$ . Let ${s_n = h_n t_n}$ and we write ${u_n =s_n w_n }$ with ${\lim s_n = \lim h_n \lim t_n =1\times 1=1}$ . Thus ${u_n \sim w_n}$

7. Let ${u_n = O\left(1/n\right)}$ and ${v_n = O (1/ \sqrt{n})}$ . Show that ${u_n v_n = o ( 1/n^{4/3})}$ .

${u_n = h_n 1/n}$ and ${v_n = t_n 1/ \sqrt{n}}$ with ${h_n}$ and ${t_n}$ bounded sequences. Now

${u_n v_n = \dfrac{h_n}{n} \dfrac{t_n}{\sqrt{n}}= \dfrac{h_n t_n}{n^{3/2}}=\dfrac{h_n t_n}{n^{1/6}}\dfrac{1}{n^{4/3}}}$

Let ${s_n = \dfrac{h_n t_n}{n^{1/6}}}$ it is ${\lim s_n = \lim \dfrac{h_n t_n}{n^{1/6}} = 0}$ since ${h_n t_n}$ is bounded.

Thus it is ${u_n v_n = o (1/n^{4/3})}$

8. Using Stirling’s approximation show that ${\log n! = n\log n -n + O(\log n)}$

We know that it is ${\log n! = n\log n -n + +r_n}$ with ${1<r_n<1+\log n}$ . Thus it follows

${\begin{array}{lcr} \dfrac{1}{\log n}<\dfrac{r_n}{\log n}<\dfrac{1+\log n}{\log n} \\ 0<\dfrac{1}{\log n}<\dfrac{r_n}{\log n}<\dfrac{1}{\log n} +1\leq \dfrac{1}{\log 2}+1 \end{array}}$

Where we used the fact that ${ \dfrac{1}{\log n}+1}$ is decreasing function. Thus ${\dfrac{r_n}{\log n}}$ is bounded and so ${r_n=O(\log n)}$ as desired.

Written by ateixeira

July 29, 2009 at 5:48 pm

Posted in Basic Mathematics, Real Analysis

Real Analysis – Limits and Continuity IV

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As an application of the previous theorem 15 let us look into the functions ${f(x)=e^x}$ and ${g(x)=\log x}$ . Now ${f:\mathbb{R} \rightarrow \mathbb{R^+}}$ and is a strictly increasing function, and ${g:\mathbb{R^+} \rightarrow \mathbb{R}}$ also is a strictly increasing function.

By the previous theorem it is $\lim_{x \rightarrow +\infty} \exp x= \mathrm{sup} [\mathbb{R^+}] = +\infty$ and $\lim_{x \rightarrow -\infty}\exp x = \mathrm{inf} [\mathbb{R^+}] = 0$ .

As for ${g(x)}$ we have $\lim_{x \rightarrow +\infty} \log x = \mathrm{sup} [\mathbb{R}] = +\infty$ and $\lim_{x \rightarrow 0} \log x = \mathrm{inf} [\mathbb{R}] = -\infty$ .

Definition 16 Let ${D \subset \mathbb{R}}$ ; ${f,g: D \rightarrow \mathbb{R}}$ , and ${c \in D'}$ .Let us suppose that there exists ${h: D \rightarrow \mathbb{R}}$ such as ${f(x) = h(x)g(x) }$ .

If $\lim_{x \rightarrow c} h(x)=1$ we say that ${f(x)}$ is asymptotically equal to ${g(x)}$ when ${x \rightarrow c}$ and write ${f(x) \sim g(x)\,\, (x \rightarrow c)}$ .
If $\lim_{x \rightarrow c} h(x) = 0$ we say that ${f(x)}$ is little-o of ${g(x)}$ when ${x \rightarrow c}$ and write ${ f(x) = o (g(x)) \,\, (x \rightarrow c)}$ .
If ${h(x)}$ is bounded in some neighborhood of ${c}$ we say that ${f(x)}$ is big-o of ${g(x)}$ when ${x \rightarrow c}$ and write ${f(x)=O(g(x)) \,\, (x \rightarrow c)}$ .

If it is the case that in the previous definition ${g(x)}$ doesn’t equal zero, we have the following equivalences:

${ f(x) \sim g(x) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 1}$ .
${ f(x) = o (g(x)) \,\, (x \rightarrow c) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 0}$ .
${ f(x) = O(g(x)) \,\, (x \rightarrow c) \Leftrightarrow \dfrac{f(x)}{g(x)} }$ is bounded in some neighborhood of ${c}$ .

This notions work exactly as they worked for sequences and they give the same type of information about the behavior of the functions in question.

Theorem 17 Let ${D \subset \mathbb{R}}$ ; ${f,g,f_0,g_0: D \rightarrow \mathbb{R}}$ , and ${c \in D'}$ . Then:

If ${f(x) \sim g(x) \,\, (x \rightarrow c)}$ and $\lim_{x \rightarrow c}g(x) = a$ , then $\lim_{x \rightarrow c} f(x) = a$
If ${f(x) \sim f_0(x) \,\, (x \rightarrow c)}$ and ${g(x) \sim g_0(x) \,\, (x \rightarrow c)}$ , then ${f(x)g(x) \sim f_0(x)g_0(x) \,\, (x \rightarrow c)}$ and ${f(x)/g(x) \sim f_0(x)/f_0(x) \,\, (x \rightarrow c)}$ .

Proof:

A proof of this simple theorem is left for the reader as an exercise

$\Box$

As an example of the previous definitions we can say, with full generality, that for any polynomial function we can keep track of the term with the leading degree if we are interested in how it behaves for larger and larger values.

But on the other hand if we are interested on how the polynomial function behaves near the origin we have to keep track of the term with the smaller degree. To see that this is indeed so let us introduce the following example:

$\displaystyle f(x) = x^2+x$

Now ${x^2+x=(x+1)x}$ . If we take ${h(x)=x+1}$ it is $\lim_{x \rightarrow 0} h(x)=1$ and so it is ${x^2+x \sim x \,\, (x \rightarrow 0)}$ .

Pay special attention to the previous example cause I’ve lost count of the number of times I see people keep the leading term in a polynomial function near the origin and get the answers all wrong.

Another example that has a lot of interest to us is:

$\displaystyle \sin x \sim x \,\, (x \rightarrow 0)$

We can see that it is so because of $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$

— Epsilon-delta condition —

And it is time for us to introduce the concept of limit using the much loved and talked about ${ \epsilon - \delta }$ condition. Once again we are walking into regions of greater and greater rigor (so that we are more certain of what we say) at the expense of having to use more abstract concepts while we are doing it. Things are going to get a little harder for people that aren’t used to this types of reasoning but please bear with me and you’ll find it rewarding when you get used to it.

The point of the ${ \epsilon - \delta }$ is for us to avoid using fuzzy concepts as near, input signals, output signals, or the somewhat weak definition of limit we been using so far.

Theorem 18 (Heine’s Theorem) Let ${D \subset \mathbb{R}}$ , ${f: D \rightarrow \mathbb{R}}$ , ${c \in D'}$ and ${a \in \overline{\mathbb{R}}}$ . We have that $\lim_{x \rightarrow c} f(x) = a$ if and only if

$\displaystyle \forall \delta > 0 \, \exists \epsilon >0 : \,\, x \in V(c,\epsilon) \cap (D \setminus \left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a, \delta)$

Proof:

Omitted.

$\Box$

In case you are wondering what that means the straightforward answer is that it means exactly what you’re idea of a function having a limit in a given point is (I’m assuming you have the right idea). It tell us that if a function indeed have limit ${a}$ in point ${c}$ that if we restrict ourselves to points near ${c}$ than the images of those points are all near ${a}$ .

Once again I tell the reader to look at this as if it were a game played between two (slightly odd) people. One of them is choosing the ${\delta}$ and the the other is choosing the ${\varepsilon}$ . But this game isn’t just about choosing.The first player gets to choose any ${\delta}$ he wants but the second has to choose the right ${\varepsilon}$ that makes the condition hold. If he can prove that he has an ${\varepsilon}$ for every ${\delta}$ that the other player chooses than he succeeds in the game and the function does have limit ${a}$ at point ${c}$ .

Theorem 19 Let ${D \subset \mathbb{R}}$ , ${f: D \rightarrow \mathbb{R}}$ , and ${c \in D'}$ . If $\lim_{x \rightarrow c} f(x)$ exists and is finite, than there exists a neighborhood of ${c }$ where ${f(x)}$ is bounded.

Proof:

Let $\lim_{x \rightarrow c} f(x) = a \in \mathbb{R}$ . By theorem 18 with ${\delta=1}$ there exists ${\varepsilon > 0}$ such as

$\displaystyle x \in V(c,\varepsilon)\cap(D\setminus\left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a,1) \Rightarrow$

$\Rightarrow f(x) \in \left] a-1, a+1\right[$

Thus $x \in V(c, \varepsilon) \cap (D \setminus \left\lbrace c \right\rbrace) \Rightarrow a-1 < f(x) < a+1$

So ${x \in V(c,\varepsilon) \cap D \Rightarrow f (x) \begin{cases} \leq \mathrm{max} \left\lbrace a+1,f(c)\right\rbrace \\ \geq \mathrm{max}\left\lbrace a+1,f(c)\right\rbrace \end{cases} }$

and ${f(x)}$ is bounded in ${V(c,\varepsilon)}$

$\Box$

— Application —

If $\lim_{x \rightarrow c} f(x)/g(x)$ exists, then ${f(x)= O(g(x))\,\, (x \rightarrow c)}$ since in this case it is ${h(x)=f(x)/g(x)}$ and there exists some neighborhood of ${c}$ where ${h(x)}$ is bounded.

After this one may be interested in knowing how we can translate $\lim_{x \rightarrow c^+} f(x) = a$ to a ${\varepsilon - \delta}$ condition. In this case we are considering ${f(x)}$ only in the set ${D_{c^+}}$ and so what we get is:

$\displaystyle \forall \delta > 0 \exists \varepsilon > 0: \, x \in V(c,\varepsilon)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta)$

Theorem 20 Let ${D \subset \mathbb{R}}$ , ${f:D \rightarrow \mathbb{R}}$ , and ${c \in D'}$ . If $\lim_{x \rightarrow c^-}f(x)=\lim_{x \rightarrow c^+}f(x)=a$ , then $\lim_{x \rightarrow c}f(x)=a$ .

Proof:

Let ${\delta>0}$ . By the ${\varepsilon-\delta}$ condition it is:

$\displaystyle \exists \varepsilon_1>0:x \in V(c,\varepsilon_1)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta)$

$\displaystyle \exists \varepsilon_2>0:x \in V(c,\varepsilon_2)\cap D_{c^-} \Rightarrow f(x) \in V(a,\delta)$

Thus by taking ${\varepsilon =\mathrm{min} \left\lbrace \varepsilon_1, \varepsilon_2 \right\rbrace }$ it follows ${x \in V(c,\varepsilon) \cap (D \setminus \left\lbrace c \right\rbrace ) \Rightarrow x \in V(c,\varepsilon) \cap D_{c^+}}$ or ${x \in V(c,\varepsilon) \cap D_{c^- }\Rightarrow f(x) \in V(a,\delta)}$

In conclusion:

${ \forall \delta>0 \exists \varepsilon >0: x \in V(c,\varepsilon)\cap (D\setminus \left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a,\delta) }$ which is equivalent to saying that $\lim_{x \rightarrow c} f(x)=a$

$\Box$

Definition 21Let ${D \subset \mathbb{R}}$ ; ${f: D \rightarrow \mathbb{R}}$ and ${c \in D}$ . We say that ${f(x)}$ is continuous in point ${c}$ if for all sequences ${x_n}$ of points in ${D}$ , such as ${\lim x_n = c}$ we have ${\lim f(x_n)=f(c)}$ .

A function is said to be continuous if it is continuous in all points in ${D}$ .

— Applications —

$\displaystyle f(x)=|x| \quad \forall x \in \mathbb{R}$

Let ${c \in \mathbb{R}}$ and ${x_n}$ a sequence such as ${x_n \rightarrow c}$ . Then ${f(x_n)=|x_n|}$ and ${\lim f(x_n) = \lim |x_n| = |c|}$ . In conclusion ${f(x_n) \rightarrow f(c)}$ which is equivalent to saying that ${f}$ is continuous in ${c}$ . Since ${c}$ can be any given point ${f(x)=|x|}$ is continuous in ${\mathbb{R}}$ .
Let ${f(x)= \sin x}$ and ${x_n}$ a sequence such as ${x_n \rightarrow \theta}$ . It is ${\lim \sin x= \sin \theta}$ and by the same reasoning ${\sin x}$ is also continuous.
In general if ${x_n \rightarrow c}$ it is ${\lim f(x_n)=f(c)=f(\lim x_n)}$ . So for ${\exp (x)}$ we have ${\lim \exp (x_n)=\exp (\lim x_n)}$ If ${x_n \rightarrow +\infty }$ it follows that ${\lim \exp(x_n)=+\infty }$ and for ${x_n \rightarrow -\infty}$ it follows that ${\lim \exp(x_n)=0}$ . So if we define ${\exp (+\infty)=+\infty}$ and ${\exp (-\infty)=0}$ it follows that it always is ${\lim \exp (x_n)=\exp (\lim x_n)}$ .
Analogously we can define ${\log +\infty= +\infty}$ and ${\log 0 = -+\infty}$ and it always is ${\lim \log x_n = \log (\lim x_n)}$ .

Theorem 22 (Heine’s theorem for continuity) Let ${D \subset \mathbb{R}}$ , ${f:D \rightarrow \mathbb{R}}$ and ${c \in D}$ . ${f}$ is continuous in ${D}$ if and only if

$\displaystyle \forall \delta>0 \,\,\exists \, \varepsilon > 0: \, x \in D \wedge |x-c|<\varepsilon \Rightarrow |f(x)-f(c)|<\delta$

Or written in terms of neighborhoods

$\displaystyle \forall \delta>0 \,\,\exists \, \varepsilon > 0: \, x \in V(c,\varepsilon) \cap D \Rightarrow f(x) \in V(f(c),\delta)$

Proof: Omitted. $\Box$

As can be seen the ${\varepsilon - \delta}$ condition for continuity in point ${c}$ is very similar to the one for limit ${a}$ in point ${c}$ . This fact causes people to sometimes confuse both concepts. And this confusion isn’t all that alarming because both concepts are indeed related. But we can see both concepts as a kind of a measure of how well behaved a function is. I think that one can look as the concept of continuity being a more stringent one in what concerns good behavior but I rather look at them as two concepts that complete and complement each other.

To finish this post I’ll just state a theorem that sheds some light on the connections of these two concepts:

Theorem 23 Let ${D \subset \mathbb{R}}$ , ${f:D \rightarrow \mathbb{R}}$ and ${c \in D \cap D'}$ . Then ${f}$ is continuous in point ${c}$ if and only if $\lim_{x \rightarrow c} f(x) = c$

Proof: Omitted. $\Box$

So as this theorem shows the connection between continuity and limit is indeed a deep one, but we can look at the concept of limit as being an auxiliary tool to determine if a function is continuous or not and we should not confuse them. In the next post I intend to write a little bit more about continuity but in the mean time a very good text about it can be found here

Written by ateixeira

July 1, 2009 at 2:25 pm

Posted in Basic Mathematics, Real Analysis

Announcement

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Due to severe lack of time the blog will only be updated in a month or so. Until then feel free to ask for clarifications or give any feedback that you find pertinent.

See you all soon.

Written by ateixeira

June 5, 2009 at 10:32 pm

Posted in Uncategorized

Real Analysis – Limits and Continuity III

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As promised in the last post I’ll start by explaining a little bit more carefully what we are trying to convey with the formalization of the concept of limit.

The first thing I want to say is that the concept of limit is a local one. In mathematical lingo what this means is that for a function to have a limit in a given point, $\lim_{x \rightarrow c} f(x) = a$ , it doesn’t matter how the function behaves when we are far away from the point in question, what matters is just how the function behaves in the vicinity of the point. This all very good for common day to day knowledge but it is not good enough for Mathematics. In Mathematics we want to be the most rigorous and formal so that very few doubts are left in the end (this is an oversimplified since this issue alone could be the focus of a blog (or a book)). So, with the concept of limit what we are doing is formalizing what do we mean with the expressions far away, vicinity.

For an example let me introduce the function

$\displaystyle f(x) = \begin{cases} o \quad x \in \mathbb{Q}\\ x \quad x \in \mathbb{R}\setminus \mathbb{Q} \end{cases}$

This is not a very sophisticated function, but it serves for what I’m trying to convey. First of all let us plot this function to see what it looks like.

onelimitfunction

Where we have drawn the ${x \in \mathbb{Q}}$ case in blue and the ${x \in \mathbb{R}\setminus \mathbb{Q}}$ case in red.

It is easy to see that for all ${c}$ different than ${0}$ the function has no limit. For ${c \neq 0}$ $\lim_{x \in \mathbb{Q} \rightarrow c} f(x) = 0$ and ${ \displaystyle\lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x) = c }$ . So $\lim_{x \in \mathbb{Q} \rightarrow c} f(x) \neq \lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x)$ , thus we can conclude that this limit doesn’t exist. For ${c=0}$ it is possible to prove (I’ll do that when the concept of limit is formalized using an ${ \epsilon-\delta }$ condition) that $\lim_{x \rightarrow 0} f(x) = 0$ .

They don’t make concepts more local than this! This function only has a limit at point ${0}$ . In an intuitive way we can understand this result like this: we can interpret the concept of limit as a measure of how good behaved a function is. Since this function is always jumping from point to point as we move from rational numbers to irrational numbers we can say that it isn’t a well behaved one. Well, the former statement is true almost everywhere in the domain of the function. The only point where it breaks down is at point ${0}$ . This is so because even though the function is a badly behaved one it misbehaves less and less while ${x \rightarrow 0}$ .

Now getting back into our normal course we’ll continue generalizing the theorems we proved for sequences into the real functions:

Theorem 12 Let ${D \subset \mathbb{R} }$ , ${f,g : D \rightarrow \mathbb{R}}$ , ${c \in D'}$ ; and let us suppose that there exists ${r>0}$ such as ${f(x) \leq g(x)\quad \forall x \in V(c,r) \cap \left( D \setminus \left\lbrace c\right\rbrace \right) }$ . Then, if $\lim_{x \rightarrow c} f(x)= +\infty$ it also is $\lim_{x \rightarrow c} f(x)= +\infty$ . And if $\lim_{x \rightarrow c} g(x)= -\infty$ it also is $\lim_{x \rightarrow c} f(x)= -\infty$

Proof:

Omitted.

$\Box$

The previous theorem states a very straightforward fact, but nevertheless, as always, what matters is that this result can be proven. In more prosaic terms this theorem expresses the conditions that need to fulfilled for us to know the limits of some functions just by knowing the limit of another one. It may well be the case that one limit may be very easy to calculate while the other is not. But, if we can establish an order relationship and calculate one of the limits it is possible for us to conclude something about the limit of the other function. In Theorem 12 we where particularly interested in the cases when the limit is ${ \pm \infty}$ but we already seen in Theorem 10 that limit weakens order relationships. In this case if we have ${f(x)\leq g(x)}$ , for some neighborhood around a point ${c}$ , then we know that $\lim_{x \rightarrow c} f(x)\leq \lim_{x \rightarrow c} g(x)$ also. Now, if ${\displaystyle\lim_{x \rightarrow c}f(x)=+\infty}$ ${g(x)}$ has no choice but to go to positive infinity as we move closer to ${c}$ since it has to be larger than ${f(x)}$ . In the case $\lim_{x \rightarrow c} g(x) = -\infty$ a similar reasoning applies. ${f(x)}$ is smaller than ${g(x)}$ and if ${g(x)}$ gets to smaller and smaller values as we approach ${c}$ than ${f(x)}$ also has to get smaller and smaller values.

Theorem 13 (Squeezed function theorem) Let ${D \subset \mathbb{R} }$ , ${f,g : D \rightarrow \mathbb{R}}$ , ${c \in D'}$ ; and let us suppose that there exists ${r>0}$ such as ${g(x) \leq f(x) \leq h(x)\quad \forall x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace }$ . Then, if $\lim_{x \rightarrow c} g(x) = \lim_{x \rightarrow c} h(x) = a$ it also is $\lim_{x \rightarrow c} f(x) = a$ .

Proof:

Omitted.

$\Box$

This theorem continues the trend of computing limits of functions without computing them! In here if we can box the function in a neighborhood of a point by two functions, and if we compute the limits of the boxing functions and come to the conclusion that they are equal we are able to know that the boxed function has the same limit.

As an example let us see the limit

$\displaystyle \lim_{x \rightarrow +\infty} \frac{\sin x}{x}$

We have that ${-1 \leq \sin x \leq 1 \quad \forall x \in \mathbb{R}}$ . Thus $-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \quad \forall x > 0$

Since $\lim_{x \rightarrow +\infty}-\frac{1}{x}=\lim_{x \rightarrow +\infty}\frac{1}{x}= 0$ it also is $\lim_{x \rightarrow +\infty}\frac{\sin x}{x}=0$ .

squeezedsin

As a second example let us now look into:

$\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}$

We have $\cos x < \frac{\sin x}{x} < 1\quad \forall x \in \left] -\frac{\pi}{2},0 \right[ \cup \left] 0,\frac{\pi}{2}\right[$

It is $\lim_{x \rightarrow 0}1=1$ and $\lim_{x \rightarrow 0} \cos x = \cos 0 = 1$ . Thus it also is $\lim_{x \rightarrow 0} \frac{\sin x}{x}=1$

squeezedsinorigin

Theorem 14 (Algebraic properties of limits) Let ${D \subset \mathbb{R}}$ ; ${f,g:D \rightarrow \mathbb{R}}$ and ${c \in D'}$ . Then:

$\lim_{x \rightarrow c} f(x)=a \Rightarrow \lim_{x \rightarrow c} |f(x)|=|a|$
$\lim_{x \rightarrow c} f(x)=a$ and $\lim_{x \rightarrow c} g(x)=b$ , then $\lim_{x \rightarrow c} \left( f(x)+g(x)\right) = a+b$
If $\lim_{x \rightarrow c} f(x) = +\infty$ and ${g}$ bounded below, than $\lim_{x \rightarrow c} (f(x)+g(x))= +\infty$
If $\lim_{x \rightarrow c} f(x) = -\infty$ and ${g}$ bounded above, than $\lim_{x \rightarrow c} (f(x)+g(x))= -\infty$
If $\lim_{x \rightarrow c} f(x) = 0$ and ${g}$ bounded, than $\lim_{x \rightarrow c} (f(x)g(x))= 0$
If $\lim_{x \rightarrow c} f(x) = a$ and $\lim_{x \rightarrow c} g(x) = b$ , than $\lim_{x \rightarrow c} (f(x)g(x))= ab$
If $\lim_{x \rightarrow c} f(x) = +\infty$ and $\lim_{x \rightarrow c} g(x) = a \neq 0$ , than $\lim_{x \rightarrow c} |f(x)g(x)|= +\infty$
If $\lim_{x \rightarrow c} f(x) = a \neq 0$ , than $\lim_{x \rightarrow c} 1/f(x)= 1/a$
If $\lim_{x \rightarrow c} f(x) = +\infty$ , than $\lim_{x \rightarrow c} 1/f(x)= 0$
If $\lim_{x \rightarrow c} f(x) = 0$ , than $\lim_{x \rightarrow c} 1/|f(x)|= +\infty$

Proof:

We’ll only prove the second one since the reasoning is mostly the same for all propositions.

Let ${x_n}$ be a sequence in ${D \setminus \left\lbrace c \right\rbrace }$ such as ${x_n \rightarrow c}$ . Then ${f(x_n) \rightarrow a}$ and ${g(x_n) \rightarrow b}$ . And from what we already saw for sequences it is ${f(x_n)+g(x_n) \rightarrow a+b}$ . By definition of limit it is $\lim_{x \rightarrow c} (f(x)+g(x)) = a + b$ .

$\Box$

Theorem 15 (Monotone function Theorem) Let ${D \subset \mathbb{R}}$ ; ${f: D \rightarrow \mathbb{R}}$ , ${ \alpha = \mathrm{inf}D}$ and ${ \beta = \mathrm{sup} D}$ .

Then:

If ${ \alpha \in D' }$ , $\lim_{x \rightarrow \alpha} f(x)$ exists and it is: $\lim_{x \rightarrow \alpha} f(x) = \mathrm{inf}f \left[ D_{\alpha^+} \right]$ if ${f}$ is increasing. $\lim_{x \rightarrow \alpha} f(x) = \mathrm{sup}f \left[ D_{\alpha^+} \right]$ if ${f}$ is decreasing.
If ${ \beta \in D' }$ , $\lim_{x \rightarrow \beta} f(x)$ exists and it is: $\lim_{x \rightarrow \alpha} f(x) = \mathrm{sup}f \left[ D_{\beta^-} \right]$ if ${f}$ is increasing. $\lim_{x \rightarrow \alpha} f(x) = \mathrm{inf}f \left[ D_{\beta^-} \right]$ if ${f}$ is decreasing.

Proof:

A formal proof of this theorem won’t be given but I’ll provide a plot of a function to help us visualize this theorem. Here I don’t think that the proof is all that important, but what counts is the intuition behind the result.

As an example of an increasing function I give ${f(x) = \sin x \quad \forall x \in \left] -\pi/2, \pi/2\right[ }$

increasing_function

In this case it is ${ \alpha = -\pi/2 }$ and ${ \beta = \pi/2 }$ ; $\lim_{x \rightarrow -\pi/2} \sin x = \sin(-\pi/2)= -1$ . ${ D_{\alpha^+} }$ represents ${ D \cap \left] \alpha, +\infty \right[ }$ So that ${f \left[ D_{\alpha^+} \right] }$ represents the image of ${f}$ by ${ D \cap \left] \alpha, +\infty \right[ }$ that is to say that ${f \left[ D_{\alpha^+} \right] = \left] -1, 1 \right[ }$ and ${ \mathrm{inf}\left] -1, 1 \right[=-1 }$ as we already seen when calculating the limit.

In a similar way we can also check that it indeed is $\lim_{x \rightarrow \pi/2} \sin x = \sin(\pi/2)= f \left[ D_{\beta^-} \right]$

For the decreasing function, $f(x)= \cos x \quad \forall x \in ]0,\pi[$ both steps are to be done by the reader.

decreasing_function

$\Box$

And this is where we stop the current post. In the next one I’ll give an application of this theorem and will introduce the ${ \epsilon-\delta }$ reasoning.

Written by ateixeira

April 28, 2009 at 12:41 pm

Posted in Basic Mathematics, Real Analysis

Real Analysis – Limits and Continuity II

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Example 4

$\displaystyle \lim_{x \rightarrow 0^+} \frac{1}{x}$

In this case it is ${D_{0^+} = \left] 0, +\infty\right[ }$ and ${ 0^+ \in D_{0^+} }$ . If ${x_n}$ is a sequence of points in ${D_{0^+}}$ such as ${x_n \rightarrow 0^+}$ it follows ${ \lim f(x_n) = \lim \dfrac{1}{x_n} = \dfrac{1}{0^+} = + \infty }$

After this simple example we’ll introduce a theorem that will state a somewhat obvious result. But we have to put our intuitions on a firm ground and this is exactly what this theorem will do. In layman terms what it expresses is that if a function has a limit in a given point ${c}$ than the one-sided limits have to be equal and equal to the limit of the function. In a more kinematic way it tell us if we approach ${c}$ by points of the domain either always to the right of ${c}$ , or always to the left of ${c}$ , that values of the images of those points have to converge to the same value.

Theorem 9 Let ${D \subset \mathbb{R}}$ , ${f: D \rightarrow \mathbb{R} }$ , ${c \in D'}$ and let us suppose that $\lim_{x \rightarrow c} f(x) = a$ . Then, if ${c \in D'_{c^+}}$ it also is $\lim_{x \rightarrow c^+} f(x) = a$ ; and if ${c \in D'_{c^-}}$ it also is $\lim_{x \rightarrow c^-} f(x) = a$ .

Proof:

Let ${x_n}$ be a sequence of points in ${D_{c^+} }$ such as ${x_n \rightarrow c}$ . Since ${x_n}$ is a sequence of points in ${D \setminus \left\lbrace c \right\rbrace}$ (by our choice of ${x_n}$ ) and $\lim_{x \rightarrow c} f(x)=a$ (by the hypothesis of the theorem) it follows from the definition of limit that ${ \lim f(x_n) = a }$ . But this is just $\lim_{x \rightarrow c^+} f(x) = a$ by Definition 8.

The case $\lim_{x \rightarrow c^-} f(x)$ is proved in the same way with the due modifications and is left as an exercise for the reader.

$\Box$

Example 5

$\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x}$

It is easy to see that this limit doesn’t exist using the previous axiom. We already know that $\lim_{x \rightarrow 0^+} \dfrac{1}{x} = + \infty$ and that $\lim_{x \rightarrow 0^-} \dfrac{1}{x} = - \infty$ . Since the limit from the right of ${0}$ is different from the limit of the left of ${0}$ we can conclude that this limit doesn’t exist.

Just in case the previous example has caused some doubts on the reader I’ll now try to explain in a more clear way the reasoning behind it. We can say, in a pretty relaxed way, that we used a reasoning by contradiction. Theorem 9 is an implication theorem. By that I mean a theorem that states a relationship between two propositions where the fact of one them being true implies that the other one is also true.

So, we have proposition ${P}$ and proposition ${Q}$ . And an implication between those two propositions can be for instance: “The validity of ${Q}$ depends on the validity of ${P}$ “. What this means is that if ${P}$ is a true statement than the validity of ${Q}$ will follow.And that if ${Q}$ is a false statement than ${P}$ will also be a false statement. In mathematical notation: (where ${\neg P}$ means not ${P}$ ) ${P \Rightarrow Q \Leftrightarrow \neg Q \Rightarrow \neg P}$ Note that if we have ${\neg P}$ we can’t conclude anything about the logical value of ${Q}$ and that if we have ${Q}$ we can’t conclude anything about the value of ${P}$ . An everyday situation may helps us here:

Imagine that you are waiting for and old friend from an uncle of yours called Pierre. You have never known Pierre and the only thing that you know about him is that he only speaks French. So a fellow comes to you and starts asking for directions in English. At that moment you can conclude that the fellow in question isn’t Pierre ( ${ \neg Q \Rightarrow \neg P }$ ). If by chance some fellow comes near you speaking French than you can’t conclude anything (remember that Pierre isn’t the only French speaking guy in Planet Earth).

In Theorem 9 we had $\lim_{x \rightarrow c} f(x) = a \Rightarrow \lim_{x \rightarrow c^+} f(x) = a \land \lim_{x \rightarrow c^-} f(x) = a$ . In this case ${P}$ is $\lim_{x \rightarrow c} f(x) = a$ and Q is $\lim_{x \rightarrow c^+} f(x) = a \land \lim_{x \rightarrow c^-} f(x) = a$ . So by showing that $\lim_{x \rightarrow 0^+} \dfrac{1}{x} \neq \lim_{x \rightarrow 0^-} \dfrac{1}{x}$ we arrived at the conclusion we have ${\neg Q}$ and so ${\neg P}$ has to follow. In this case ${\neg P}$ is just the statement that $\lim_{x \rightarrow 0}\dfrac{1}{x} = a$ is meaningless statement for any ${a}$ and so $\lim_{x \rightarrow 0}\dfrac{1}{x}$ doesn’t exist.

We will now state a group of theorems that generalize what we already saw for sequences.

Theorem 10 (Limit of Inequalities) Let ${D \subset \mathbb{R}}$ , ${f,g : D \rightarrow \mathbb{R}}$ , ${c \in D'}$ and let us suppose that there exists ${r>0}$ such as ${ f(x)<g(x)\quad \forall x \in V(c,r) \cap (D\setminus \left\lbrace c \right\rbrace ) }$ . If $\displaystyle \lim_{x \to c} f(x)$ and $\displaystyle \lim_{x \to c} g(x)$ exist it is $\displaystyle \lim_{x \to c} f(x) \leq \lim_{x \to c} g(x)$
Proof:

Let ${x_n}$ be a sequence of points in ${D \setminus \left\lbrace c \right\rbrace }$ such as ${x_n \rightarrow c}$ . By the definition of limit of a sequence ${\exists k \in \mathbb{N}:\quad n \geq k \Rightarrow x_n \in V(c,r) \Rightarrow x_n \in V(c,r) \cap D\setminus \left\lbrace c \right\rbrace }$ .

Since ${x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace \Rightarrow f(x) \leq g(x)}$ . So ${n \geq k}$ implies that ${f(x_n) \leq g(x_n)}$ . By Theorem 1.4 we know that it is $\lim f(x_n) \leq \lim g(x_n)$ . Since $\lim_{x \rightarrow c} f(x) = f(x_n)$ and $\lim_{x \rightarrow c} g(x) = g(x_n)$ it follows $\lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} g(x)$

$\Box$

Corollary 11 Let ${D \subset \mathbb{R}}$ , ${f: D \rightarrow \mathbb{R} }$ , ${c \in D'}$ and ${a \in \mathbb{R}}$ . If there exists ${r > 0}$ such as ${f(x) \leq a}$ ( ${f(x) \geq a}$ ) ${ \forall x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace }$ and if $\lim_{x \rightarrow c} f(x)$ exist. It is $\lim_{x \rightarrow c} f(x) \leq a$ ( $\lim_{x \rightarrow c} f(x) \geq a$ ).

Proof: Take ${g(x)=a}$ in the previous theorem. $\Box$

In the next post I’ll start by discussing a little bit more the concept of limit for a real function, and then a few more theorems that generalize what we already saw for sequences will stated.

Written by ateixeira

April 15, 2009 at 9:18 pm

Posted in Basic Mathematics, Real Analysis

Climbing the Mountain

Real Analysis Exercises III

Real Analysis – Limits and Continuity IV

Announcement

Real Analysis – Limits and Continuity III

Real Analysis – Limits and Continuity II

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