# Proof of $f(x) = (e^x-1)/x = 1 \text{ as } x\to 0$ using epsilon-delta definition of a limit

I am in calc 1 and we have just learned the epsilon-delta definition of a limit and I (on my own) wanted to try and use this methodology in order to prove $(e^x-1)/x = 1$ (one of the equivalencies), along with $\displaystyle \frac {\sin(x)}{x} = 1$, that the proof just told us "was so."

I do not know how to put the happy little math symbols in this website so I'm going to upload a picture of my work. Now, I understand how to apply the epsilon-delta definition of the limit for some easy problems, even for some complex functions where the numbers simply "fall out," but what do I do with the the $|f(x)-L|<\epsilon$ after I've made it be $|(e^x-1-x)/x| < \epsilon$?

I understand that I basically need to get $|(e^x-1-x)/x|$ to become equivalent to $|x|$ but how do I do this? Is this factorable?

And if this kind of easy problem is difficult for me, does this mean that I do have what it takes to become a math major? I really love this kind of problem-solving but sometimes I just don't get the answer. Thanks!

http://tinypic.com/r/wiae6f/7

The above is my problem.

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What is your definition for $e^x$? Without this you cannot expect to solve the problem. Also, different starting definitions will lead to different solutions. – Eric Naslund Jun 1 '11 at 23:55
I hope that you can see from the answers that your difficulty has absolutely nothing to do with whether you have what it takes to be a math major. "I really love this kind of problem-solving" does answer the question of whether you have what it takes. – André Nicolas Jun 2 '11 at 1:32

Another approach, harder to handle rigorously than any of the ones suggested so far, is to do it the way Euler did, essentially by defining $e$ as $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n.$$ (But then we would in particular want to prove that the limit exists, which is not easy.)

Now imagine that $n$ is large, and let $h=1/n$. Then $e^h$ should be about $1+1/n$, and the rest follows. But the details, such as making precise the weaselly "should be about $1+1/n$," are not easy. In particular, we would have to define precisely the general exponential function.

So unless we fill in a lot of detail, the above idea involves quite vigorous hand waving, diametrically opposite to the epsilon-delta approach. However, the idea has useful intuitive content.

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Suppose that somehow, we know the derivative of $e^x$ is itself, that is $\frac{d}{dx}e^x=e^x$. (This could follow from the power series definition)

Then, we have that by the definition of the derivative $$e^x=\lim_{h=\rightarrow 0} \frac{e^{x+h}-e^x}{h}$$ and after dividing by $e^x$ we get $$\lim_{h=\rightarrow 0} \frac{e^{h}-1}{h}=1.$$

Again it really depends on the definition you are starting from.

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It would seem that this begs the question. At the very least, it's dissatisfying to appeal to the derivative to answer a question totally primitive to these concepts. (Even given the power series for $e^x$, we should't know anything about the derivative). – barf Jun 2 '11 at 5:08

Are you allowed to use $e^x = 1 + x + x^2/2! + \cdots$? If so, then show that $1+x \leq e^x \leq 1 + x + x^2$ for all $x$ in a neighborhood of $0$.

EDIT (elaborating): Assuming the definition $e^x = \sum\nolimits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}$, you can show that $$1+x \leq e^x \leq 1 + x + x^2$$ holds for all $x$ in some $\delta$-neighborhood of $0$, very simply as follows. On the one hand, $$e^x -1 - x = x^2\bigg(\frac{1}{{2!}} + \frac{x}{{3!}} + \frac{{x^2 }}{{4!}} + \cdots \bigg),$$ from which the first inequality is immediately seen to hold; on the other hand, $$e^x -1 - x - x^2 = -x^2 \bigg(\frac{1}{{2!}} - \frac{x}{{3!}} - \frac{{x^2 }}{{4!}} - \cdots \bigg),$$ from which the second inequality is immediately seen to hold. Indeed, note that for any $r > 0$ (as small as we wish), it holds $$\sup _{|x| \le r} \Big(\Big|\frac{x}{{3!}}\Big| + \Big|\frac{{x^2 }}{{4!}}\Big| + \cdots \Big) = \frac{r}{{3!}} + \frac{{r^2 }}{{4!}} + \cdots \le r + r^2 + \cdots = \frac{r}{{1 - r}}.$$

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Considering they haven't been taught how to take a derivative, the power series expansion of an analytic function is likely illegal. – syxiao Jun 2 '11 at 0:20
Apparently, this is the most common definition. – Shai Covo Jun 2 '11 at 0:24
@syxiao: Why? You don't have to know derivatives in order to define $e^x = \sum\nolimits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}$. – Shai Covo Jun 2 '11 at 0:28
@Shai, if all you have is that definition, how do you prove $e^ae^b=e^{a+b}$? That is, how do you justify the notation $e^x$? – Gerry Myerson Jun 2 '11 at 0:34
@Shai, yes, all usual properties of the exponential functions can be shown starting with the power series --- but think about trying to do that with a class that has barely learned what a limit is! You can't even assume they have any idea what an infinite series is, what convergence means, how the ratio test goes, let alone the manipulative skills needed to go from the power series to the basic facts about exponentials. – Gerry Myerson Jun 2 '11 at 0:58

I suggest changing the limit of $x$ tending to zero to some function of $x$ tending to infinity thru a mathematically manipulation. For example if we use $1/h = e^x-1$ then $e^x = 1+1/h$.

Remembering we want to find $\lim_{x\to 0} (e^x-1)/x$. If now we use log to help with our manipulation $\ln (e^x)= \ln(1+ 1/h)$

now as $x$ tending to zero $h$ tending to infinity as it is essentially has a inverse effect

now substituting into our original limit we get

$\lim_{h\to \infty}{1/h/\ln(1+1/h)} = 1/\ln(1+1/h)^h)$ and since by definition $(1+1/h)^h = e then taking logs of both sides \ln(e) = 1$

therefore $\lim_{h\to \infty}(1/1)$ is simply one, therefore $\lim_{x\to0} (e^x-1)/x$ must also be one. QED

I'm no math whiz and this may not withstand rigorous mathematical proof but hopefully it is a step in the right direction Cheers

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Something very odd in the line that says $e=\ln(e)$. – Gerry Myerson Oct 16 '12 at 12:25
Hello Gerry u r rite e does not equal ln(e) the equal sign should not be there but instead it should read therefore ln(e) equals 1 – leo Oct 16 '12 at 19:01
So ... edit it! – Gerry Myerson Oct 16 '12 at 21:50
Gerry it has now been edited thanks for your observation Cheers Leo – leo Oct 16 '12 at 23:22
leo, better have another look. – Gerry Myerson Oct 17 '12 at 0:12

When I teach these topics, I note that $(2^x-1)/x$ seems to be about $0.7$ for small values of $x$, while $(3^x-1)/x$ seems to be about $1.1$ for small values of $x$, so it stands to reason that there's some number between $2$ and $3$, let's call it $e$, such that $(e^x-1)/x$ tends to $1$ as $x\to0$. Not very rigorous, I know, and no deltas-and-epsilons, but as others have mentioned you have to start with $\it some$ definition of $e$ to even ask the question.

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This is the standard way of defining $e$ in most Calculus books, and IMO from a mathematical point of view is soo wrong... Basically we reason the following way, usually long before defining the derivative: Since the function $f(x)=a^x$ is differentiable at $0$, and it`s derivative is continuous as a function of $a$, we an define the number $e$. And the funny part is that most of the times we use this fact later to prove that $f(x)=a^x$ is differentiable (which for the exponential function is equivalent to differentiability at $0$).... – N. S. Jun 2 '11 at 11:01
That's essentially equivalent to Euler's definition of $e$, by the way :) – Alexei Averchenko Oct 16 '12 at 11:54

The only way I see this question being attackable is to know what the derivative of $e^x$ is. In many ways $e^x$ is defined by this property. In this case I can offer two solutions.

1) Suppose you know that $\frac{d}{dx} e^x = e^x$. Then we prove that $1 + x \leq e^x \leq 1 + x + x^2$ for $x$ sufficiently close to 0. To see this, define $f(x) = e^x - 1 - x$. Since $f(0) = 0$, it suffices to show that $f(0)$ is a local minimum. If $x \geq 0$, then $f'(x) = e^x - 1$, which is positive, so $f$ is increasing for $x \geq 0$. On the other hand if $x < 0$, then $f'$ is negative, so $f$ is decreasing to the left of zero. Hence $0$ is a local minimum for $f$ and so $f$ is positive for a small neighborhood around 0. Now, let $g(x) = e^x - 1 - x - x^2$. Again, $g(0) = 0$. It will now suffice to show that 0 is a local maximum for $g$. To see this, we have $g'(x) = e^x - 1 - 2x$. It is not so clear whether this is positive or negative for $x$ close to 0, so we take the derivative again to get $g''(x) = e^x - 2$, which is negative for $x < ln 2$. Hence $g$ is concave down in a neighborhood around 0, which implies that 0 is a local maximum. Hence we have $g(x) \leq 0$ in a neighborhood around 0. Combining the results for $f,g$ we get $1 + x \leq e^x \leq 1 + x + x^2$. This yields that $\lim_{x \rightarrow 0} x/x \leq \lim_{x \rightarrow 0} (e^x - 1)/x \leq \lim_{x \rightarrow 0} (x + x^2)/x$, so by the squeeze theorem the desired limit is 1.

2) If you know L'Hospital's Rule, then the result follows immediately since $\lim_{x \rightarrow 0} (e^x - 1)/x = \lim_{x \rightarrow 0} e^x/1 = e^0 = 1$.

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It is worth mentioning that for l'Hopitals rule you also need to know the derivative of $e^x$. A bit of care is needed. For example to use L'Hopitals rule to prove that $\frac{\sin(x)}{x}\rightarrow 1$ as $x\rightarrow 0$ is not always feasible (unless we start from power series for $\sin(x)$) because we need to know the derivative of $\sin(x)$, and to take the derivative of in the first place you usually use that limit. – Eric Naslund Jun 2 '11 at 0:43

I don't know if you also wanted information on $\lim_{x\to0}\sin x/x$, but the standard way to make that one plausible is with a diagram. Draw the circle of radius $1$ centered at the origin, locate the point $P=(\cos x,\sin x)$, then $\sin x$ is the distance from $P$ to the $x$-axis if you go straight down, while $x$ is the distance from $P$ to the $x$-axis if you go along the circle (I guess I'm assuming $P$ is in the first quadrant). It is plausible that as $x$ approaches zero the ratio of these two lengths approaches $1$.

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