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I want to evaluate the following limit:

$$\lim_{x\rightarrow0}\frac{e^x-e^{-x}}{e^x-1}$$

For $x\rightarrow0$, the denominator is asymptotic to

$$e^x-1\sim x$$

Here's how I simplify the numerator:

$$e^x-e^{-x}\sim 1-e^{-x}=-(e^{-x}-1)\sim-(-x)=x$$

Finally we have

$$f(x)\sim\frac{x}{x}=1$$

My textbook does it another way and the solution is 2. I wonder what I'm doing wrong. Perhaps it's the fact that I have replaced $e^x$ with $1$ but that seemed to make sense to me because $e^x\rightarrow1$ for $x\rightarrow0$. Any hints?

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  • $\begingroup$ Try L'Hospital's Rule !! $\endgroup$ – Anik Bhowmick Aug 12 '18 at 9:19
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How do you justify $e^x-e^{-x}\sim 1-e^{-x}$?

It is true that $e^x = 1 + o(1)$, but if that is your justification then $\sim$ must mean "up to terms of degree $1$ or more", so when you end up with the numerator being $\sim x$, then this is the same as $\sim 0$ or $\sim 2x$ as far as this meaning of $\sim$ is concerned. You've thrown away the information you need to conclude anything.

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What you did is wrong because both $e^x$ and $e^{-x}$ behave as $1$ near $0$. You can't replace one if them by $1$, while the other one remains as it is.

Note that$$\lim_{x\to0}\frac{e^x-e^{-x}}{e^x-1}=\lim_{x\to0}\frac{e^x+e^{-x}}{e^x}=2.$$

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The solution goes as follows: $$\lim_{x\to 0}\frac{e^x-e^{-x}}{e^x-1}=\lim_{x\to 0}\frac{\frac{e^{2x}-1}{e^x}}{e^x-1}=\lim_{x\to 0}\frac{\frac{(e^x-1)(e^x+1)}{e^x}}{e^x-1}=\lim_{x\to 0}\frac{e^x+1}{e^x}=2$$

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  • $\begingroup$ The OP states: "My textbook does it another way and the solution is 2". They already have a solution that leads to the correct answer, and don't need to be shown yet another one. The question was went wrong in their attempt, and this post doesn't answer that. $\endgroup$ – Henning Makholm Aug 12 '18 at 10:58
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$\displaystyle\lim_{x \to 0}\dfrac{e^x-e^{-x}}{e^x-1}$

now, $e^x-e^{-x}=\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots \right)-\left(1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\cdots \right)=2\left(x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots \right) \approx2x$ as $ x\to 0$

and $e^x-1=x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots \approx x $, as $ x\to 0$

$\therefore \displaystyle\lim_{x \to 0}\dfrac{e^x-e^{-x}}{e^x-1}=\displaystyle\lim_{x \to 0}\dfrac{2x}{x}=2.$

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You cannot just substitute the limits, as that leads to $0/0$. L'Hopital's rule is designed for exactly this type of problems, see: https://en.wikipedia.org/wiki/L%27Hôpital%27s_rule

And indeed, the answer is $2/1=2$, once you apply that rule.

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Using equivalents as you did (which the simplest method here), observe the numerator is $$\mathrm e^x-\mathrm e^{-x}=2\sinh x\sim_0 2x,$$ whence $$\frac{\mathrm e^x-\mathrm e^{-x}}{\mathrm e^x-1}\sim_0\frac{2x}x=2.$$

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Making $y = e^x$ we have

$$\lim_{x\rightarrow0}\frac{e^x-e^{-x}}{e^x-1}\equiv \lim_{y\to 1}\frac{y-\frac 1y}{y-1} = \lim_{y\to 1}\frac{y^2-1}{y(y-1)} = \lim_{y\to 1}\frac{y+1}{y}= 2 $$

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