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Compute $\displaystyle \lim_{x\to0}{1 - e^{-x}\over e^x - 1}$.

So first and foremost, I know what the answer to this question is:

$$\begin{align}\lim_{x\to0}{1 - e^{-x}\over e^x - 1} &= \lim_{x\to0}\left({1 - e^{-x}\over e^x - 1}\right)\cdot{e^x\over e^x} \tag{$\star$}\\ & = \lim_{x\to0}{e^x - 1\over (e^x - 1)e^x} \\ &= \lim_{x\to0}{1 \over e^x} \\ &= {1\over e^0} \\ &= 1.\end{align}$$

Unfortunately, I found this solution by complete accident. I can't really justify the motivation of why I performed the step $(\star)$ other than "I saw the $e^{-x}$ and thought to try and get rid of it.

Initially the first thought that I had was to use the conjugate of one of the functions, but we end up not going anywhere with the problem (There's still that nasty $e^{-x}$ using either conjugate). Is this likely the intended solution to this problem? I unfortunately cannot see any other way of solving the problem.

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  • $\begingroup$ You could use L'Hospitals rule at the beginning to reduce to the answer! $\endgroup$ – Zach Sep 19 '18 at 22:44
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    $\begingroup$ Since you selected the tag "limits-without-lhopital", I presume that comments or answers instructing you to use L'Hopital aren't what you're looking for? $\endgroup$ – Namaste Sep 19 '18 at 22:48
  • $\begingroup$ @amWhy that's correct I would assume $\endgroup$ – Ahmad Bazzi Sep 19 '18 at 22:49
  • $\begingroup$ It's not clear to me whether you are asking how you can know why, in this case, multiplying by $\frac {e^x}{e^x}$ was a good idea? (It is a good idea). Or are you looking for alternative methods? $\endgroup$ – Namaste Sep 19 '18 at 22:53
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    $\begingroup$ @amWhy, yes. A student asked me how to compute this limit, and unfortunately I could not come up with a better answer than presented here. The student in question is an introductory calculus student, and derivatives have not been covered yet. Hence, I wanted to know if there was something else that I might be missing in regards to this problem that makes it simpler. Incidentally, it appeared in the section where conjugates are discussed, and thus is the 1st step I went for in explaining the problem. But when that didn't work, I faithlessly tried multiplying by $e^x/e^x$ and it ended up working $\endgroup$ – Decaf-Math Sep 19 '18 at 22:54
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Here's three ways of doing it:

Taylor

You can use Taylor, i.e. $e^{x} \sim 1 + x$, so $$\frac{1 - e^{-x}}{e^{x} - 1} \sim \frac{1 - (1-x)}{1+x-1} = \frac{x}{x} =1$$

Factorizing

Another (but equivalent) way is factor $e^{x}$ from the denominator $$\displaystyle \lim_{x\to0}{1 - e^{-x}\over e^x - 1} = \displaystyle \lim_{x\to0}{1 - e^{-x}\over e^{x}(1 - e^{-x})}= \lim_{x\to0}\frac{1}{e^{x}} = \frac{1}{e^0} = 1 $$

L'Hopital

You can solve by L'Hopital, i.e. $$\displaystyle \lim_{x\to0}{1 - e^{-x}\over e^x - 1} = \displaystyle \lim_{x\to0} \frac{e^{-x}}{e^{x}} = \frac{e^{0}}{e^{0}} = 1$$

You could use $(\epsilon,\delta)$ definition as well

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  • $\begingroup$ You could also define $f(x)=\frac{1-e^{-x}}{e^{x}-1}$, which has the property that $f(x)=f(-x)^{-1}$. Take limits as $x \rightarrow 0$ on both sides, and note that the limit must be positive. ($f(x)=f(-x)^{-1}$ implies that if the right one-sided limit exists and is not zero, so does the left one. Existence of right limit follows directly from continuity of $f$ on $\mathbb{R}\setminus \left \{ 0 \right \}$) $\endgroup$ – LPenguin Sep 19 '18 at 23:39
  • $\begingroup$ That's right @LPenguin you could do that as well $\endgroup$ – Ahmad Bazzi Sep 19 '18 at 23:42
  • $\begingroup$ Why does existence of the right limit follow from continuity? It could as easily be the case that $f(x)$ blows up as $x\to0$... $\endgroup$ – Steven Stadnicki Sep 19 '18 at 23:46
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You may also use the substitution $x=\ln y$. This may make your trick seem more natural: $${1 - e^{-x}\over e^x - 1} = {1 - e^{-\ln y}\over e^{\ln y} - 1} ={1 - {1 \over y}\over y - 1} ={y - 1\over y(y - 1)} = { 1\over y}\stackrel{y \to 1}{\longrightarrow}1$$

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  • $\begingroup$ Another awesome way!!! $\endgroup$ – DarkKnight Sep 20 '18 at 2:35
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Another way is writing $e^{-x}$ as $\dfrac{1}{e^x}$.

Then, $$\begin{align}\lim_{x\to0}{1 - e^{-x}\over e^x - 1} &= \lim_{x\to0}\left({1 - \dfrac{1}{e^{x}}\over e^x - 1}\right) \\ & = \lim_{x\to0}{e^x - 1\over (e^x - 1)e^x} \\ &= \lim_{x\to0}{1 \over e^x} \\ &= {1\over e^0} \\ &= 1\end{align}$$

See? Almost same as your solution. I think one can easily see through these steps.

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My own approach would be to notice the similarity between the numerator and denominator, and by educated guess perform the factorization

$$e^x-1=e^x(1-e^{-x})$$ and the rest is easy.

Or to perform the substitution $t:=e^x$, giving

$$\frac{1-\dfrac1t}{t-1}=\frac1t.$$

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