Real Analysis – Limits and Continuity VI

— More properties of continuous functions —

Definition 35

Let {D \subset \mathbb{R}}; {f: D\rightarrow \mathbb{R}} and {c \in D'\setminus D}. If {\displaystyle \lim_{x\rightarrow c}f(x)=a\in \mathbb{R}}, we can define {\tilde{f}} as:

\displaystyle   \tilde{f}(x)=\begin{cases} f(x) \quad x \in D \\ a \quad x=c \end{cases} \ \ \ \ \ (16)

As an application of the previous definition let us look into {f(x)= \sin x/x}. It is {D= \mathbb{R}\setminus \{0\}}.

Since {\displaystyle\lim_{x \rightarrow 0} \sin x/x=1} we can define {\tilde{f}} as

\displaystyle  \tilde{f}(x)=\begin{cases} \sin x/x \quad x \neq 0 \\ 1 \quad x=0 \end{cases}

As another example let us look into {f(x)=1/x} Since {\displaystyle\lim_{x\rightarrow 0^+}f(x)=+\infty} and {\displaystyle\lim_{x\rightarrow 0^-}f(x)=-\infty} we can’t define {\tilde{f}} for {1/x}.

Finally if we let {f(x)=1/x^2} we have {\displaystyle\lim_{x\rightarrow 0^+}f(x)=\displaystyle\lim_{x\rightarrow 0^-}f(x)=+\infty}. Since the limits are divergent we still can’t define {\tilde{f}}.

In general one can say that given {f: D\rightarrow \mathbb{R}} and {c \in D'\setminus D} {\tilde{f}} exists if and only if {\displaystyle\lim_{x \rightarrow c}f(x)} exists and is finite.

Theorem 42 Let {D \subset \mathbb{R}}; {f,g: D\rightarrow \mathbb{R}} and {c \in D}. If {f} and {g} are continuous functions then {f+g}, {fg} and (if {g(c)\neq 0}){f/g} are also continuous functions.

Proof: We’ll prove that {fg} is continuous and let the other cases for the reader.

Let {x_n} be a sequence of points in {D} such that {x_n \rightarrow c}. Then {f(x_n) \rightarrow f(c)} and {g(x_n) \rightarrow c} (since {f} and {g} are continuous functions).

Hence it follows {f(x_n)g(x_n) \rightarrow f(x)g(x)} from property {6} of Theorem 19. Which is the definition of a continuous function. \Box

Let {f(x)=5x^2-2x+4}. First we note that {f_1(x)=5}, {f_2(x)=-2} and {f_3(x)=4} are continuous functions. Now {f_4(x)=4} also a continuous function. {f_5(x)=x^2} is continuous since it is the product of {2} continuous functions. {f_6(x)=-2x} is continuous since it is the product of {2} continuous functions. Finally {f(x)=5x^2-2x+4} is continuous since it is the sum of continuous functions.

Theorem 43 Let {D, E \subset \mathbb{R}}, {g: D\rightarrow E}, {f: E \rightarrow \mathbb{R}} and {c \in D}. If {g} is continuous in {c} and {f} is continuous in {g(c)}, then the composite function {f \circ g (x)=f(g(x)) } is continuous in point {c}.

Proof: Let {x_n} be a sequence of points in {D} with {x_n \rightarrow c}. Hence {\lim g(x_n)=g(c)}. If {f} is continuous in {g(c)} it also is {\lim f(g(x_n))=f(g(c))}. This is {\lim (f \circ g)(x_n)= (f \circ g)(c)}. Thus {f \circ g} is continuous in {c}. \Box

As an application of the previous theorem let {f(x)=a^x}. Since {a^x=e^{\log a^x}=e^{x \log a}} we can write {a^x=e^t \circ t=x\log a}. Since {f(t)=e^t} is a continuous function and {g(x)=x \log a} is also a continuous function it follows that {a^x} is a continuous function (it is the composition of two continuous functions).

By the same argument we can also show that with {\alpha \in \mathbb{R}}, {x^\alpha} (for {x \in \mathbb{R}^+}) is also a continuous function in {\mathbb{R}^+}.

Theorem 44 Let {D, E \subset \mathbb{R}}, {g: D\rightarrow E}, {f: E \rightarrow \mathbb{R}} and {c \in D'}. Suppose that {\displaystyle \lim_{x \rightarrow c}g(x)=a} and that {\displaystyle \lim_{t \rightarrow a}f(t)} exists. If {f} is continuous it follows {\displaystyle \lim_{x \rightarrow c}f(g(x))=\lim_{t \rightarrow a}f(t)}.

Proof: Omitted. \Box

Find {\displaystyle \lim_{x \rightarrow +\infty} \sin (1/x)}.

We can write {\sin (1/x)= \sin t \circ (t=1/x)}. Since {\displaystyle \lim_{x \rightarrow + \infty}(1/x)=0} it is, from Theorem 44, {\displaystyle \lim_{x \rightarrow +\infty} \sin (1/x)=\displaystyle\lim_{t \rightarrow 0}\sin t =0}.

In general if {\displaystyle \lim_{x \rightarrow c} g(x)= a \in \mathbb{R}} it is {\displaystyle \lim_{x \rightarrow c} \sin (g(x))=\displaystyle\lim_{t \rightarrow a} \sin t = \sin a}. In conclusion

\displaystyle  \lim_{x \rightarrow c}\sin (g(x))=\sin (\lim_{x \rightarrow c}g(x))

Suppose that {\displaystyle \lim_{x \rightarrow c}g(x)=0} and let {\tilde{f}} be the function that makes {\sin x/x} be continuous in {x=0}.

It is {\sin x = \tilde{f}(x)x}, hence it is {\sin g(x) = \tilde{f}(g(x))g(x)}.

By definition {\tilde{f}} is continuous so by Theorem 44 {\displaystyle \lim_{x \rightarrow c^+}f(g(x))=\displaystyle\lim_{t \rightarrow 0}\tilde{f}(t)=1}.

Thus we can conclude that when {\displaystyle \lim_{x \rightarrow c}g(x)=0} it is

\displaystyle  \sin (g(x))\sim g(x)\quad (x \rightarrow c)

For example {\sin (x^2-1) \sim (x^2-1)\quad (x \rightarrow 1)}.

Let {\displaystyle \lim_{x \rightarrow c}g(x)=a \in \mathbb{R}}. By Theorem 44 it is {\displaystyle \lim_{x \rightarrow c} e^{g(x)}=\lim_{t \rightarrow a}e^t=e^a} (with the conventions {e^{+\infty}=+\infty} and {e^{-\infty}=0}). Thus {\displaystyle \lim_{x \rightarrow c}e^{g(x)}=e^{\displaystyle\lim_{x \rightarrow c}g(x)}}.

Analogously one can show that {\displaystyle \lim_{x \rightarrow c} \log g(x)= \log (\lim_{x \rightarrow c}g(x))} (with the conventions {\displaystyle \lim_{x \rightarrow +\infty} \log g(x)=+\infty} and {\displaystyle \lim_{x \rightarrow 0} \log g(x)=-\infty}).

Let {a>1}. It is {\displaystyle \lim_{x \rightarrow +\infty}a^x =\displaystyle\lim_{x \rightarrow +\infty}e^{x\log a}=e^{\displaystyle\lim_{x \rightarrow +\infty} x\log a}=+\infty } (since {\log a>0}).

On the other hand for {\alpha > 0} it also is {\displaystyle \lim_{x \rightarrow +\infty}x^\alpha =\displaystyle\lim_{x \rightarrow +\infty}e^{\alpha \log x}= e^{\displaystyle \lim_{x \rightarrow +\infty}\alpha \log x}=+\infty}.

The question we want to answer is {\displaystyle \lim_{x \rightarrow +\infty}\dfrac{a^x}{x^\alpha} } since the answer to this question tell us which of the functions tends more rapidly to {+\infty}.

Theorem 45 Let { a<1} and {\alpha > 0}. Then

\displaystyle   \lim_{\infty}\frac{a^x}{x^\alpha}=+\infty \ \ \ \ \ (17)

Proof: Let {b=a^{1/(2\alpha)}} ({b>1}). It is {a=b^{2\alpha}}. Hence {a^x=b^{2\alpha x}}. Moreover {\dfrac{a^x}{x^\alpha}=\dfrac{b^{2\alpha x}}{x^\alpha}=\dfrac{b^{2\alpha x}}{\sqrt{x}^{2\alpha}}}.

which is

\displaystyle   \frac{a^x}{x^\alpha}=\left( \frac{b^x}{\sqrt{x}} \right)^{2\alpha} \ \ \ \ \ (18)

Let {[x]} denote the nearest integer function and using Bernoulli’s Inequality ({b^m\geq 1+ m(b-1)}) it is {b^x\geq x^{}[x]\geq 1+[x](b-1)>[x](b-1)>(x-1)(b-1)}.

Hence {\dfrac{b^x}{\sqrt{x}}>\dfrac{x-1}{\sqrt{x}}(b-1)=\left( \sqrt{x}-1/\sqrt{x}\right)(b-1)}.

Since {\displaystyle \lim_{x \rightarrow +\infty}\left( \sqrt{x}-1/\sqrt{x}\right)(b-1)=+\infty} it follows from Theorem 32 that {\displaystyle\lim_{x \rightarrow \infty} \frac{b^x}{\sqrt{x}}=+\infty}.

Using 18 and setting {t=b^x/\sqrt{x}} it is {\displaystyle\lim_{x \rightarrow \infty}\frac{a^x}{x^\alpha}=\displaystyle\lim_{t \rightarrow +\infty}t^{2\alpha}=+\infty} \Box

Corollary 46 Let {\alpha > 0}, then

\displaystyle \lim_{x \rightarrow +\infty}\frac{x^\alpha}{\log x}=+\infty

Proof: Left as an exercise for the reader (remember to make the convenient change of variable). \Box

Theorem 47 Let {a>1}, then {\displaystyle \lim \frac{a^n}{n!}}=0. Proof: First remember that {\log n!=n\log n -n + O(\log n)} which is Stirling’s Approximation.

Since {\dfrac{\log n}{n} \rightarrow 0} it also is {\dfrac{O(\log n)}{n} \rightarrow 0}.


\displaystyle \frac{a^n}{n!}=e^{\log (a^n/n!)}=e^{n\log a - \log n!}


\displaystyle \lim \frac{a^n}{n!}=e^{\lim(n\log a - \log n!)}

For the argument of the exponential function it is

{\begin{aligned} \lim(n\log a - \log n!) &= \lim n\log a-n\log n+n-O(\log n) \\ &=\lim \left(n\left(\log a -\log n+1 -\dfrac{O(\log n)}{n}\right)\right) \\ &=+\infty\times -\infty=-\infty \end{aligned}}

Hence {\displaystyle \lim \frac{a^n}{n!}=e^{-\infty}=0}. \Box

Lemma 48

\displaystyle   \lim_{x \rightarrow +\infty}\left( 1+\frac{1}{x}\right)^x=e \ \ \ \ \ (19)

Proof: Omitted. \Box

Theorem 49

\displaystyle   \lim_{x \rightarrow 0}\frac{\log (1+x)}{x}=1 \ \ \ \ \ (20)

Proof: Will be proven as an exercise. \Box

Corollary 50

\displaystyle   \lim_{x \rightarrow 0}\frac{e^x-1}{x} \ \ \ \ \ (21)

Proof: Left as an exercise for the reader. Make the change of variables {e^x=t+1} and use the previous theorem. \Box

Generalizing the previous results one can write with full generality:

  • {\sin g(x) \sim g(x) \quad (x \rightarrow c)} if {\displaystyle \lim_{x \rightarrow c} g(x)=0}
  • {\log (1+g(x)) \sim g(x) \quad (x \rightarrow c)} if {\displaystyle \lim_{x \rightarrow c} g(x)=0}
  • {e^{g(x)}-1 \sim g(x) \quad (x \rightarrow c)} if {\displaystyle \lim_{x \rightarrow c} g(x)=0}

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