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how would you show that $$\lim_{x\to 0}\frac{e^x-1}{x}=1$$ without using derivatives or l'hopital but using basic ideas that are generally introduced just before derivatives in a typical introductory calculus course. my attempt.. $$\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\lim_{n\to \infty}\frac{\left(1+\frac{x}{n}\right)^n-1}{x}$$ then.. this is

$$ =\lim_{x\to 0}\lim_{n\to \infty}\frac{1+{n \choose 1}\frac{x}{n} + {n \choose 2}\left(\frac{x}{n}\right)^2 +\cdots+ {n \choose n}\left(\frac{x}{n}\right)^n -1}{x}$$

$$ =\lim_{x\to 0}\lim_{n\to \infty} \left[ 1+ \frac{ {n \choose 2}\left(\frac{x}{n}\right)^2 +\cdots+ {n \choose n}\left(\frac{x}{n}\right)^n }{x}\right]$$

then we could switch the limit order but the justification for this seems to be not very basic and I would like to provide a basic explanation, as basic as possible.

by the way I am doing this to provide a pre-derivatives proof of $$\lim_{x\to c}\frac{e^x-e^c}{x-c}=e^c$$

thank you.

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Since $e^x$ is "transcendental," any way we define it we're going to run into issues with swapping the order of limits. So we'll have to use some polynomial functions to do the dirty work for us. The only polynomials on hand come from the interest formula definition $$ e^x = \lim_{n\to\infty} \bigg( 1 + \frac{x}{n}\bigg)^n $$ so let's see what we can do with those.

First, show that $1+x < e^x$ for all $x$. This is going to be the "less than" function in an application of the squeeze theorem.

Now the thing to do is take the second-order Taylor polynomial and check what happens. (We're not allowed to talk about "Taylor polynomials," technically, but for $n$ large the binomial expansion of the interest formula is basically the same.)

There are two qualitatively distinct cases. If $x > 0$, then the interest rate formula binomial expansion (i.e. Taylor series) has terms all positive, so the second-order polynomial $1 + x + x^2/2$ is less than the exponential. No good. We'll come back to this in a second.

If $x<0$, then the second order approximation majorizes the exponential function. We know this has to be true because of the alternating series error theorem from Taylor series, but since you're not allowed to use that, let's restrict to $-1<x<0$ and notice that $|x^n|<|x^k|$ if $n>k$, so definitely $|x^n|/n! < |x^k|<k!$. That's the idea in the alternating series error theorem, applied to this particular case.

Back to $x>0$. Let's "cheat" again by observing that there's a whole pencil of parabolas which pass through $(0,1)$ and have $1+x$ as their tangent line at $0$. If we choose one with a large enough quadratic coefficient, we should be set to use the squeeze theorem. Because $1$ is a nice coefficient, let's see if $1+x+x^2$ works. We need $1+x+x^2 > e^x,$ and again I'll indicate the argument with the Taylor series, trusting that you can convert to the binomial approximation.

$$ 1+x+x^2 > 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots$$ is the same as $$ \frac{1}{2}x^2 > \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$ i.e. $$ \frac{1}{2} > \frac{x}{3!} + \frac{x^2}{4!} + \cdots.$$

We're limiting at zero, so we can let $x$ be small and positive: $|x|<1$, say. In this case, $$ \frac{1}{2} > \frac{x}{3!} + \frac{x^2}{4!} + \cdots > x\bigg(\frac{1}{3!} + \frac{1}{4!} + \cdots\bigg).$$ That last converges to $e-2$, in fact (this will be easier if you're using binomial approximation I think), so we see that for $0 < x < \frac{1}{2(e-2)}$, we have that $e^x < 1 + x + x^2$.

Bonus: $1+x+x^2 > 1 + x + x^2/2$ for all $x$, so ... for $|x|<\frac{1}{2(e-2)}$, $$ 1 + x \leq e^x \leq 1 + x + x^2 $$ so $$ \frac{(1+x)-1}{x} \leq \frac{e^x-1}{x} \leq \frac{(1+x+x^2)-1}{x} $$ and by the squeeze theorem you have the desired result.


Remarks.

  • When using the binomial expansion, you need to show that these inequalities hold for the interest rate formula with any (sufficiently large) $n$. It shouldn't be too bad because the coefficients of the binomial series are limiting to the coefficients of the Taylor series.

  • To make rigorous the use of Taylor series, we need to appeal to the uniform convergence of power series in closed intervals. The general fact is this: Suppose $f_n$ is a sequence of functions and $x$ is in their domain. If $f_n$ converges uniformly, you can swap the limits: $$\lim_{t\to x}\lim_{n\to\infty} f_n(t) = \lim_{n\to\infty}\lim_{t\to x}f_n(t).$$ This and more is in chapter 7 of Rudin. Results about power series, and the exponential function in particular, are in chapter 8.

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  • $\begingroup$ Certainly brilliant, but I'm afraid it is not very usual to define $e$ this way in elementary teaching.Instead, almost every book begins with $$e\overset{def}{=}\displaystyle\lim_{n\to\infty}(1+\frac1n)^n$$ $\endgroup$ – Vim Feb 23 '15 at 5:42
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    $\begingroup$ @Vim If I'm not mistaken, you can use your definition to derive Neal's expression for $e^x$ for rational $x$ (by taking a subsequence of the your sequence $e^x = (1+\frac{1}{n})^{nx}$ so that $nx$ is integer, and seeing that this is also a subsequence of the monotonically increasing sequence $(1+\frac{x}{n})^n$), and then argue by continuity to extend to irrational $x$. This is perhaps less 'simple', but it does use the more standard definition. $\endgroup$ – Strants Feb 23 '15 at 5:57
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    $\begingroup$ How do you prove the exponential function is differentiable without using the limit that we want to compute, which —mind you— is precisely the one which computes its derivative at zero? $\endgroup$ – Mariano Suárez-Álvarez Feb 24 '15 at 5:52
  • $\begingroup$ @MarianoSuárez-Alvarez Is that question directed at me? $\endgroup$ – Neal Feb 24 '15 at 21:16
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Showing this$$\lim_{x\to 0^+}\frac{e^x-1}{x}=1$$ amounts to showing: $$\lim_{n\to\infty}\frac{e^{\frac{1}{n}}-1}{\frac1n}=1$$where $n\in\mathbb N^+$
Now take a look at this $$\lim_{n\to\infty}\frac{\frac1n}{\frac1n}=\lim_{n\to\infty}\frac{[(1+\frac1n)^n]^{\frac1n}-1}{\frac1n}=1$$ which is of course obvious.
Note that $(1+\frac1n)^{n}$ is increasing as $n$ goes up and its limit is $e$, therefore $$\frac{e^{\frac{1}{n}}-1}{\frac1n}>\frac{[(1+\frac1n)^n]^{\frac1n}-1}{\frac1n}\tag{1}$$ Then we go on and look at this $$\lim_{n\to\infty}\frac{[(1+\frac{1}{n-1})^{n}]^{\frac1n}-1}{\frac1n}=\lim_{n\to\infty}\frac{\frac{1}{n-1}}{\frac1n}=1$$ And note that $(1+\frac{1}{n-1})^{n}$ decreases as $n$ goes up and its limit is $e$ as well, so $$\frac{[(1+\frac{1}{n-1})^{n}]^{\frac1n}-1}{\frac1n}>\frac{e^{\frac{1}{n}}-1}{\frac1n}\tag{2}$$ Put $(1),(2)$ together, squeeze, and we're done :)


EDIT
For the $x \to 0^-$ part, you can start with the fact that $$\lim_{n\to -\infty}(1+\frac1n)^n=e$$ The rest to do are very alike, except for some adjustments because $n$ is negative now.
Or, if you hate dealing with a negative $n$, it doesn't matter to just replace $n$ with $-n$ and let $n$ still tend to $+\infty$. For example, rewrite $\lim_{n\to -\infty}(1+\frac1n)^n=e$ as $\lim_{n\to\infty}(1-\frac1n)^{-n}=e$ (which I'm sure you are more familiar with). Such a rewriting keeps $n$ still positibe and therefore will probably remove a lot of trouble when you are applying the Bernoulli Inequality in the Remark below.


Rm:
On reasons why $(1+\frac1n)^n$ is monotonically increasing (which is not trivial but a little tricky), see here for a good proof (by @Thomas) based on Bernoulli's Inequality.

Inspired by this proof, I have also made out a similar proof to show that $(1+\frac{1}{n-1})$ is monotonically decreasing. And here it goes:
Let $a_n=(1+\frac{1}{n-1})^n$ for all $n\ge 2$ and $n\in \mathbb N$. Obviously $a_n$ is always positive. Now check this for all $n$: $$a_{n}>a_{n+1}\Leftrightarrow (1+\frac1{n-1})^n > (1+\frac1n)^{n+1}\Leftrightarrow \Bigl(\frac{1+\frac{1}{n-1}}{1+\frac1n}\Bigr)^n>1+\frac1n$$ In the last equation the LHS simplifies to $(1+\frac{1}{n^2-1})^n$, according to Bernoulli $$LHS=(1+\frac{1}{n^2-1})^n>1+\frac{n}{n^2-1}>1+\frac{n}{n^2}=1+\frac1n=RHS$$ QED.

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  • $\begingroup$ Very nice! ${}$ $\endgroup$ – Neal Feb 23 '15 at 5:10
  • $\begingroup$ @Neal Thanks ${}$ $\endgroup$ – Vim Feb 23 '15 at 5:36
  • $\begingroup$ To be quite picky, you have to assume the limit exists to be justified in computing it along the $1/n$, yes? $\endgroup$ – Kevin Carlson Feb 23 '15 at 6:17
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    $\begingroup$ It is not trivial, in fact. $\endgroup$ – Mariano Suárez-Álvarez Feb 24 '15 at 5:55
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    $\begingroup$ @MarianoSuárez-Alvarez According to what my prof taught me, To show the "equivalence" of $$\lim_{x\to 0^+}\frac{e^x-1}{x}=1$$ and $$\lim_{n\to\infty}\frac{e^{\frac{1}{n}}-1}{\frac1n}=1$$ we need to use consider the three: $f([x]),f(x)$ and $f([x]+1)$ and use something like the sandwich theorem. $\endgroup$ – Vim Feb 24 '15 at 6:00
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Now the following is informal, but we could Taylor expand $e^x$ about $x=0$ to get: $$e^x=\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}\cdots=1+x+\frac{x^2}{2!}+\cdots$$ Now

$$\frac{e^x-1}{x}=\frac{-1+1+x+\frac{x^2}{2}+\cdots}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots$$ Hence $$\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots=1$$

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  • $\begingroup$ thanks @Uncountable .. I like it.. but Taylor has so many ideas build on top of derivatives. I suspect there is a clever squeeze theorem argument or something on that level... thank you anyhow $\endgroup$ – userX Feb 23 '15 at 3:38
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    $\begingroup$ The confusion in the OP seems to be the justification for switching limits. You haven't answered it, you've just brushed it under the rug the second equality of your last line. (Not that this is a bad approach, just it doesn't seem to really answer the question.) $\endgroup$ – Neal Feb 23 '15 at 3:39
  • $\begingroup$ @Neal You're right, my approach is (as I said) still informal, and to be honest, I don't know the exact criteria for being able to switch limits, let alone a formal proof for this. I feel like existence of both limits might be enough, but I'm not sure. $\endgroup$ – Uncountable Feb 23 '15 at 3:57
  • $\begingroup$ @Uncountable It's not just the existence of limits. In general, the usual sufficient condition is uniform convergence of the family of functions, which in this case is the partial sums. This is in chapter 7 of baby Rudin. $\endgroup$ – Neal Feb 23 '15 at 4:59
  • $\begingroup$ Your proof is reasonable but I suppose a proof based on basic definitions are what the OP desires. $\endgroup$ – Vim Feb 23 '15 at 5:39
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Here's a possible method, which uses no power series. Let's focus on the right-hand limit $x \to 0^+$. Since

$$\lim_{x\to0^+}\frac{e^x - 1}{x} = \lim_{x\to0^+}\frac{1-e^{-x}}{xe^{-x}} = \lim_{x\to0^+} \frac{e^{-x}-1}{-x} = \lim_{x\to0^-} \frac{e^x-1}{x}$$

we lose no generality doing so, and it will let us work out inequalities more easily later.

We have that $e^x = \lim_{n\to\infty} \left(1+\frac{x}{n+1}\right)^{n+1} = \lim_{n\to\infty} \left(1+\frac{x}{n+1}\right)^n$. Using this,

$$\begin{align*} \lim_{x\to0^+} \frac{e^x-1}{x} &= \lim_{x\to0^+}\lim_{n\to\infty} \frac{\left(1+\frac{x}{n+1}\right)^{n+1} - 1^{n+1}}{x}\\ & = \lim_{x\to0^+}\lim_{n\to\infty} \frac{(1+\frac{x}{n+1} - 1)(1 + (1+\frac{x}{n+1}) + \cdots + (1 + \frac{x}{n+1})^n)}{x}\\ & = \lim_{x\to0^+}\lim_{n\to\infty} \frac{1}{n+1} \sum_{k=0}^n a_k \\ &= \lim_{x\to0^+}\lim_{n\to\infty} \frac{n}{n+1}\frac{1}{n} \sum_{k=0}^n a_k\end{align*}$$

Where we have used the formula for the difference of $n^\mbox{th}$ powers to get the second line, and $a_k = \left(1+\frac{x}{n+1}\right)^k$. We have that $$1+\frac{k}{n}x \le a_k \le \left(1+\frac{x}{n+1}\right)^n$$ so, $$\lim_{x\to0^+}\lim_{n\to\infty}\frac{1}{n} \sum_{k=0}^n 1+\frac{k}{n}x \le \lim_{x\to0^+}\frac{e^x-1}{x} \le \lim_{x\to0^+}\lim_{n\to\infty} \frac{1}{n} \sum_{k=0}^n\left(1+\frac{x}{n+1}\right)^n$$ $$\lim_{x\to0^+}1 + \frac{n(n-1)}{2n^2}x \le \lim_{x\to0^+}\frac{e^x-1}{x} \le \lim_{x\to0^+}\lim_{n\to\infty} \left(1+\frac{x}{n+1}\right)^n$$ (where we have removed the factors $\frac{n}{n+1} \to 1$ as $n\to\infty$ for brevity.) Taking the limits as $n\to \infty$, $$\lim_{x\to0^+}1 + \frac{1}{2}x \le \lim_{x\to0^+}\frac{e^x-1}{x} \le \lim_{x\to0^+} e^x$$ and so evaluating the two limits we find $\lim_{x\to0^+}\frac{e^x-1}{x} = 1$, as desired.

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