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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.

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  • $\begingroup$ Where did $\frac{1}{2}h+1 \leq \frac{(e^h-1)}{h} \leq |h| + e^h$ come from? $\endgroup$ – Henry May 7 '18 at 0:22
  • $\begingroup$ @Henry ah sorry I should've specified, we wanted to bound above and below and we came up with those functions to do so. $\endgroup$ – rb612 May 7 '18 at 0:22
  • $\begingroup$ What is your definition of $e$? $\endgroup$ – Jeffery Opoku-Mensah May 7 '18 at 0:23
  • $\begingroup$ @JefferyOpoku-Mensah I suppose it's the traditional definition of $\lim \limits_{x\to \infty} (1+1/x)^x$. $\endgroup$ – rb612 May 7 '18 at 0:24
  • $\begingroup$ A complete proof would need to show why these are bounds $\endgroup$ – Henry May 7 '18 at 0:25
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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.

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  • $\begingroup$ 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. $\endgroup$ – rb612 May 7 '18 at 0:34
  • $\begingroup$ 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 $e^x$ $\endgroup$ – PhysicsMathsLove May 7 '18 at 0:35
  • $\begingroup$ I could show that it's equal to $\lim \limits_{x\to\infty} (1+\frac{1}{x})^x$ easily? $\endgroup$ – rb612 May 7 '18 at 0:39
  • $\begingroup$ Yes. It is a standard argument in elementary real analysis. For example, see: math.stackexchange.com/questions/637255/… $\endgroup$ – PhysicsMathsLove May 7 '18 at 0:48
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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 $h \to 0$.

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

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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.

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  • $\begingroup$ 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. $\endgroup$ – rb612 May 7 '18 at 0:41
  • $\begingroup$ My point is that in proving the inequalities you can not use the Taylor series. $\endgroup$ – Mohammad Riazi-Kermani May 7 '18 at 0:54

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