Proving the limit definition of derivative of $e^x$ using the squeeze theorem?

Question

My friend and I were working to see if we could use the squeeze theorem to prove that the derivative of $e^x$ is $e^x$.

We said that by definition, the derivative is

$$\frac{d}{dx} e^x =\lim \limits_{h\to0} \frac{e^{x+h}-e^x}{h} = \lim \limits_{h\to0} \frac{e^{x}(e^h-1)}{h} = \lim \limits_{h\to0} e^x *\lim \limits_{h\to0} \frac{(e^h-1)}{h}$$

$$\frac{1}{2}h+1 \leq \frac{(e^h-1)}{h} \leq |h| + e^h$$

$$\lim \limits_{h\to0} \frac{1}{2}h+1 = \lim \limits_{h\to0} |h| + e^h = 1$$

$$\lim \limits_{h\to0} e^x *\lim \limits_{h\to0} \frac{(e^h-1)}{h} = \lim \limits_{h\to0} e^x = e^x$$

However, this seems more simplified than any other proof we've seen, so we're wondering if there's a fault here. Neither of us have taken analysis yet, so we might be assuming something incorrect.

Where did $\frac{1}{2}h+1 \leq \frac{(e^h-1)}{h} \leq |h| + e^h$ come from? — Henry, May 07 '18 at 00:22
@Henry ah sorry I should've specified, we wanted to bound above and below and we came up with those functions to do so. — rb612, May 07 '18 at 00:22
@JefferyOpoku-Mensah I suppose it's the traditional definition of $\lim \limits_{x\to \infty} (1+1/x)^x$. — rb612, May 07 '18 at 00:24
@Henry assuming we did prove these bounds, would the proof be rigorous and correct? — rb612, May 07 '18 at 00:42
Why do you need to prove it in this way? I know proofs that circumvent this approach. You can use the chain rule and the derivative of the natural logarithm to prove this rule, and those two rules can be proven separately too. — CogitoErgoCogitoSum, May 07 '18 at 01:24

score 4 · Answer 1 · answered May 07 '18 at 01:02

Here is a pure squeeze theorem proof without using a Taylor expansion and using the definition $e = \lim_{n \to \infty}(1 +1/n)^n$.

For $h > 0$, let $n = \lfloor1/h\rfloor$. We have $n \leqslant 1/h < n+1$ and

$$\frac{n}{n+1}(n+1)(e^{1/n+1} - 1)= n(e^{1/n+1} - 1 ) \leqslant \frac{e^h -1 }{h} \leqslant (n+1)(e^{1/n} -1) = \frac{n+1}{n}n(e^{1/n}-1).$$

Since $n \to \infty$ iff $h \to 0$ we obtain the limit $1$ by the squeeze theorem if we can show that $n(e^{1/n} - 1) \to 1$.

This follows from the inequality

$$\left(1 + \frac{1}{n} \right)^n < e < \left(1 + \frac{1}{n} \right)^{n+1},$$

which implies

$$1 < n(e^{1/n}-1) < n\left[\left(1 + \frac{1}{n} \right)^{1/n}\left( 1+ \frac{1}{n}\right)-1 \right] \leqslant 1 + \frac{1}{n} + \frac{1}{n^2}.$$

Here we used Bernoullis's inequality $(1 + 1/n)^{1/n} < 1 + 1/n^2$ to obtain the far right inequality. A second application of the squeeze theorem gives us $n(e^{1/n}-1) \to 1$.

If $h <0$ and the limit is approached from the left we can use the above result and

$$\frac{e^h -1}{h} = \frac{e^{-|h|}-1}{-|h|} = e^{-|h|}\frac{e^{|h|}-1}{|h|},$$

since $e^{-|h|} \to 1$ as $h \to 0$.

The second squeeze application can be simplified with $e<(1+\tfrac{1}{n-1})^n$. — J.G., Dec 12 '20 at 13:01

score 2 · Accepted Answer · answered May 07 '18 at 00:32

2

A pretty nice way to do this is as follows:

Note that $$|e^h - 1 - h| = \left|\frac{h^2}{2} + \frac{h^3}{6} + \frac{h^4}{24} +...\right| $$ $$ = h^2 \left|\frac{1}{2} + \frac{h}{6} + \frac{h^2}{24} +...\right|\leq h^2 \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... \right) = h^2$$

So if $h\neq 0$, we have $\left|\frac{e^h - 1}{h} - 1 \right| \leq |h|$, i.e. $$\lim_{h \rightarrow 0} \left(\frac{e^h - 1}{h} - 1\right) = 0$$

Note that this tells us that $e^x$ is differentiable at the origin with derivative 1.

Now: $$\lim_{h \rightarrow 0} \left(\frac{e^{x+h} - e^x}{h}\right) = e^x \lim_{h \rightarrow 0} \frac{e^h - 1}{h} = e^x$$

as required.

answered May 07 '18 at 00:32

PhysicsMathsLove

3,142
1
21
39

Thank you, except doesn’t this require knowing the derivative of $e^x$ in order to create this Maclaurin expansion? Unless I suppose you define $e$ to be that series. – rb612 May 07 '18 at 00:34
Not if you define $e^x$ by the power series. Even if you use the sequence definition as you wrote in the comments, you can easily show this is equivalent to the series representation of $e^x$ – PhysicsMathsLove May 07 '18 at 00:35
I could show that it's equal to $\lim \limits_{x\to\infty} (1+\frac{1}{x})^x$ easily? – rb612 May 07 '18 at 00:39
Yes. It is a standard argument in elementary real analysis. For example, see: https://math.stackexchange.com/questions/637255/prove-that-the-limit-definition-of-the-exponential-function-implies-its-infinite – PhysicsMathsLove May 07 '18 at 00:48

Mohammad Riazi-Kermani · Answer 3 · 2018-05-07T00:35:24.803

1

How did you get these inequalities ?

$$\frac{1}{2}h+1 \leq \frac{(e^h-1)}{h} \leq |h| + e^h$$

You need to prove these inequalities to have a valid proof.

You can not use Taylor Series because your statement is used in Taylor series.

edited May 07 '18 at 00:35

answered May 07 '18 at 00:29

Mohammad Riazi-Kermani

68,728

Mohammad, thank you for your answer. I understand why we can't use the Taylor Series. I don't see exactly where in our proof we use the series expansion. – rb612 May 07 '18 at 00:41
My point is that in proving the inequalities you can not use the Taylor series. – Mohammad Riazi-Kermani May 07 '18 at 00:54

score 0 · Answer 4 · edited May 26 '18 at 16:26

Claim 1: $\lim\limits_{x\to 0} \dfrac{\ln(x+1)}{x}=1$

Proof: by properties of log, $\lim\limits_{x\to 0} \dfrac{\ln(x+1)}{x}=\lim\limits_{x\to 0} \ln[(x+1)^{1/x}]=\ln(e)=1$.

Claim 2: $\lim\limits_{y\to0}\dfrac{e^y-1}{y}=1$

Proof: putting $x=e^y-1$ then $\lim\limits_{y\to0}\dfrac{e^y-1}{y}=\lim\limits_{x\to0}\dfrac{x}{\ln(x+1)}=\lim\limits_{x\to0}\dfrac{1}{\dfrac{\ln(x+1)}{x}}=\dfrac{1}{1}=1$

With this, $$\lim\limits_{h\to0}\dfrac{e^{x+h}-e^x}{h}=e^x$$

Proving the limit definition of derivative of $e^x$ using the squeeze theorem?

4 Answers4

Linked