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.

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.

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.

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 $$ \sup _{|x| \le r} \Big(\Big|\frac{x}{{3!}}\Big| + \Big|\frac{{x^2 }}{{4!}}\Big| + \cdots \Big) = \frac{r}{{3!}} + \frac{{r^2 }}{{4!}} + \cdots \le r + r^2 + \cdots = \frac{r}{{1 - r}}. $$

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