Theorem 65 (Cauchy’s theorem) Let and , continuous such as . If and are differentiable in and doesn’t vanish in , there exists such as
Proof: It is since if it were would vanish in by Theorem 63. Let and define as (differentiable in and continuous in ) such as it is Thus by applying Theorem 63 in there exists such as . That is

The previous theorem is perhaps more of a lemma than a theorem per se. Because it will allows us to prove more important results. Also this result can be seen as providing a method of finding (very) local approximations to functions at a given point and as such it is the same as a Taylor expansion of first order (we’ll see what this means in futures posts).
Theorem 66 (Cauchy first limit rule) Let , and differentiable. Moreover doesn’t vanish in and .
Proof: Let . Since are continuous in and we can set . Let such as . Applying Cauchy’s Theorem 65 to each interval it is with . Then by the Squeezed Sequence Theorem 17 And by the definition of limit Thus Hence, by the definition of limit it is Analogously if is applying Cauchy’s Theorem 65 to each interval it is with . Just like in the previous steps it is From equation 34 and equation 35 it is Finally let . Let it is . From what we proved thus far it is
Hence, for this case it also is . The case can be proven in a similar way with the change of variable . 
Theorem 67 (Cauchy second limit rule) Let , and differentiable. Moreover doesn’t vanish in and . If exists it is
Proof: Left as an exercise for the reader. 
The two previous theorems are known by a variety of names on the mathematical literature and are one of the most used theorems in the practice of calculating limits.
A few examples will now be used to showcase their powers
Example 2 The functions and tend to as goes to . But which one of them tends to more strongly?

At the end of the last example we arrived once again at the type of limit where .
But the thing is that Cauchy’s first rule (and in fact the second rule too) can be used more than one time. Hence we’ll just apply it again (we’ll start from the begining again) just so we don’t lose our train of thought
As and exercise calculate
Another mathematical theorem from real analysis which is very important to Physics, in a conceptual level, is what we’ll call Lagrange’s theorem. Even though it is a theorem in Real Analysis it has a very nice interpretation in geometrical and in kinematic terms.
Theorem 68 (Lagrange’s theorem) Let and continuous. If is differentiable in there exists such as
Proof: In theorem 65 let and the result follows trivially. 
As I was saying before the statement and proof of this theorem it can be interpreted both geometrically and kinematically. The geometric interpretation states that the secant to the function in the interval has a given slope and that we can always find a tangent to the function in the interval whose slope is the same as the secant. Hence the straight lines defined by these secant and tangent are parallel.
In a kinematic sense if represents time and represents the distance travelled this result implies that if we transverse the distance in the time interval then we have an average speed which is
Since in this context can be interpreted as the being the instantaneous speed (or just speed for short) the previous result states that there exists a time instant in which your instantaneous speed is the same as you average speed for the whole time interval.
Example 3 Show that .
Proof: Let . Assume that and apply Theorem 68 to the interval . with . Then Assume now that and apply once again 68, but this time to the interval . with . Then Notice that in the we didn’t invert the sign of the inequality while multiplying by because and consequently . 
The last theorem has two important corollaries that we’ll state below.
Corollary 69 Let be an interval in and continuous. If exists and vanishes in the interior of , then is constant.
Proof: By reductio ad absurdum let us assume that is not constant. Then there exists such as and . Since is continuous in and differentiable in , by theorem 68 it is with . Hence which is absurd since it would imply that , which is contrary to our initial hypothesis. 
Corollary 70 () Let be an interval in and continuous. If exists and is positive (negative) in the interior of , then is strictly increasing (decreasing).
Proof: Let us take the case . Take such as . From theorem 68 it is with . Since it is and is strictly increasing. 
And with these results we finish the Differential Calculus part of our course in Real Analysis. The next theoretical posts of Real Analysis will dwell on the theory of numerical series.