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For the problem $\lim_{x\to 0} {e^x-1 \over x}=1$, I need to solve for x in order to solve using the delta epsilon definition. How do I go about solving for x in the equation $f(x)={e^x-1 \over x}$? Thanks.

What I have tried so far:

$${e^x-1\over x}-1 < .1$$ $${e^x-1\over x} < 1.1$$ $$e^x-1<1.1x$$ $$e^x<1+1.1x$$ $$\ln e^x<\ln(1+1.1x)$$ $$x<\ln(1+1.1x)$$

But this is obviously the incorrect way to go about solving for x. Hints are appreciated.

Edit: I realize now that I am supposed to be solving this with a calculator, and have not yet learned about the definition of e in limit form. However, I'm not even sure how to do this with a calculator because every epsilon-delta limit I have solved in the past has involved solving for x and forming an epsilon quotient that way.

For $\epsilon$ = .1, how would I do this using a calculator? Or rather, what form do I need to get $f$ in to set it up so I can?

alan
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    Do you have to use the delta-epsilon definition? Because this limit can be evaluated nicely just by using the definition of e. If you'd like, I can summarize this in a post below. – A. Thomas Yerger May 05 '14 at 04:08
  • @AlfredYerger I added what I have tried so far, but yes I need to use the delta epsilon definition for given values. In order to find the necessary $\delta$ i need to solve for x, but am not sure how to do it like this. Thanks – alan May 05 '14 at 04:09
  • As you are looking for a limit when $x$ goes to zero, you can limit your $\delta$ to any convenient range, and there is no harm if it turns out too small. You are interested in a bound, not an exact result – vonbrand May 05 '14 at 04:24

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The following is a good approximation, you can find more details here. $$e^{x} - 1 < x + \frac{x^2}{2-x } = \epsilon $$

So just solve this for $x$ and put the value of $\epsilon$ and get your bounds for $x$. $$x(2-x) + x^2 - \epsilon(2-x) = 0 \implies x = \frac{2\epsilon}{(2 +\epsilon)} $$

Putting $\epsilon = 0.1$ gives you $x=0.095$ which gives you $\displaystyle {\frac{e^{0.1}-1}{0.1}} - 1 \approx 0.049 < 0.1 = \epsilon$

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