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I was trying to calculate the limit of the following function:

$$ \lim_{x\to0} \frac{1}{\sin x} \cdot \ln \left(\frac{e^x -1}{x}\right) $$

My first thought was using L'Hopital's rule since $\Large \frac{e^x -1}{x}$ goes to 1 so the whole $\ln$ goes to 0.

But then I get another complicated expression, and finally I end up using L'Hopital's rule at least 5 times before getting an actual result.

Is there a wiser way for dealing this limit? (I mean, without using this rule?)

Thanks.

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Notice, $$\lim_{x\to 0}\frac{1}{\sin x}\cdot \ln\left(\frac{e^x-1}{x}\right)$$ $$=\lim_{x\to 0}\frac{\ln(e^x-1)-\ln(x)}{\sin x}$$ Applying L' Hospital's rule three times for $\frac 00$ form, $$=\lim_{x\to 0}\frac{\frac{e^x}{e^x-1}-\frac {1}{x}}{\cos x}$$ $$=\lim_{x\to 0}\frac{xe^x-e^x+1}{x(e^x-1)}\cdot \lim_{x\to 0}\frac{1}{\cos x}$$ $$=\lim_{x\to 0}\frac{xe^x-e^x+1}{xe^x-x}$$ $$=\lim_{x\to 0}\frac{xe^x+e^x-e^x}{xe^x+e^x-1}$$ $$=\lim_{x\to 0}\frac{xe^x}{xe^x+e^x-1}$$ $$=\lim_{x\to 0}\frac{xe^x+e^x}{xe^x+e^x+e^x}$$ $$=\lim_{x\to 0}\frac{xe^x+e^x}{xe^x+2e^x}=\frac{0+1}{0+2\cdot 1}=\color{red}{\frac 12}$$

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Using Taylor series $$e^x=1+x+\frac{x^2}{2}+O\left(x^3\right)$$ $$\frac{e^x-1}x=1+\frac{x}{2}+O\left(x^2\right)$$ $$\log\Big(\frac{e^x-1}x\Big)=\log\Big(1+\frac{x}{2}+O\left(x^2\right)\Big)=\frac{x}{2}+O\left(x^2\right)$$

I am sure that you can take it from here.

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You can use the Taylor series at $0$. At $0$ we have: $e^x\sim1+x+\frac{1}{2}x^2$. Thus $\frac{e^x-1}{x}\sim 1+\frac{1}{2}x$. Hence $\ln{\frac{e^x-1}{x}}\sim \frac{1}{2}x$.

Thus $\lim_{x\to 0}\frac{\ln{\frac{e^x-1}{x}}}{\sin{x}}=\lim_{x\to 0}\frac{\frac{1}{2}x}{\sin{x}}=\frac{1}{2}$

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Use Taylor expansion: $e^x-1=x+x^2/2+o(x^2)$. Thus \begin{align*} \frac{1}{\sin x}\log\left(\frac{e^x-1}{x}\right) &=\frac{x}{\sin x}\frac1x\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)\\ &=\frac{x}{\sin x}\frac1x\frac{\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)}{(x^2/2+o(x^2))/x}\frac{x^2/2+o(x^2)}{x}\\ &=\underbrace{\frac{x}{\sin x}\frac{\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)}{(x^2/2+o(x^2))/x}}_{\to1}\underbrace{\frac{x^2/2+o(x^2)}{x^2}}_{\to1/2}\\ \end{align*} Thus your limit goes to $1/2$.

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Using Taylor expansions is surely a good way to deal with this limit. However, with some care, we can limit the number of applications of l‘Hôpital to one. Actually two, but one application is hidden in the following theorem.

Theorem. If a function $F$ is differentiable in a neighborhood of $a$ (possibly excluding $a$), $F$ is continuous at $a$ and $\lim_{x\to a}F'(x)=l$ exists finite, then $F$ is also differentiable at $a$ and $F'(a)=l$.

The proof is a straightforward application of l’Hôpital: $$ \lim_{x\to a}\frac{F(x)-F(a)}{x-a}=\lim_{x\to a}F'(x) $$ since the limit is an indeterminate form $0/0$ because $F$ is continuous at $a$.


Consider the function $$ f(x)=\begin{cases} \dfrac{e^x-1}{x} & \text{for $x\ne0$}\\[6px] 1 & \text{for $x=0$} \end{cases} $$ It's easy to see that $f$ is continuous at $0$, because $x\mapsto e^x$ is differentiable at $0$, with derivative $1$. Also $$ f'(x)=\frac{xe^x-e^x+1}{x^2} $$ and, with a simple application of l‘Hôpital, we get $$ \lim_{x\to0}f'(x)=\lim_{x\to0}\frac{e^x+xe^x-e^x}{2x}= \lim_{x\to0}\frac{e^x}{2}=\frac{1}{2} $$ so $f$ is also differentiable at $0$ and $f'(0)=1/2$, because of the theorem above.

Now your limit is $$ \lim_{x\to0}\frac{\ln f(x)}{\sin x}= \lim_{x\to0}\frac{\ln f(x)}{x}\frac{x}{\sin x} $$ The second factor has limit $1$, so it can be disregarded, and, considering $g(x)=\log f(x)$ (which is of course defined in a neighborhood of $0$), you have, since $g(0)=0$, $$ \lim_{x\to0}\frac{g(x)}{x}=g'(0) $$ But $$ g'(x)=\frac{f'(x)}{f(x)} $$ so $$ g'(0)=\frac{1/2}{1}=\frac{1}{2} $$

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Let's try an elementary approach as follows \begin{align} L &= \lim_{x \to 0}\frac{1}{\sin x}\log\left(\frac{e^{x} - 1}{x}\right)\notag\\ &= \lim_{x \to 0}\frac{x}{\sin x}\cdot\frac{1}{x}\log\left(\frac{e^{x} - 1}{x}\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x}\log\left(\frac{e^{x} - 1}{x}\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x}\cdot\dfrac{\log\left(1 + \dfrac{e^{x} - 1 - x}{x}\right)}{\dfrac{e^{x} - 1 - x}{x}}\cdot\frac{e^{x} - 1 - x}{x}\notag\\ &= \lim_{x \to 0}\frac{1}{x}\cdot 1\cdot\frac{e^{x} - 1 - x}{x}\notag\\ &= \lim_{x \to 0}\frac{e^{x} - 1 - x}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{e^{x} - 1}{2x}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{2}\notag \end{align} We have used the standard limits $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to 0}\frac{\sin x}{x} = 1,\, \lim_{t \to 0}\frac{\log(1 + t)}{t} = 1$$ The use of L'Hospital's Rule (only once) can be avoided but at a significant cost by using the limit formula $e^{x} = \lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n}$.

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