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$\lim\limits_{x\to 0}{e^x-e^{-x} \over 1- \cos x}= \lim\limits_{x\to 0}{e^x+e^{-x} \over 1+ \sin x} = 2$

To find the value of the above lim, I had used L'Hôpital's rule since original one is a form of $0 \over 0$. Is this correct?


additional question : how to type in the letter ô ? I always copy and paste and it's inconvenient

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  • $\begingroup$ Right click -> Show Math as -> TeX commands $\hat{o}$ $\endgroup$ – alphacapture Jun 26 '17 at 17:34
  • $\begingroup$ yes it is correct $\endgroup$ – qbert Jun 26 '17 at 17:35
  • $\begingroup$ This is correct. $\endgroup$ – Crostul Jun 26 '17 at 17:35
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$$\lim\limits_{x\to 0}{e^x-e^{-x} \over 1- \cos x}=\lim\limits_{x\to 0}\frac{(e^{2x}-1)(1+\cos{x})} {e^x(1- \cos^2 x)}=\lim\limits_{x\to 0}\frac{2(e^{2x}-1)} {\sin^2x}=$$ $$=4\lim\limits_{x\to 0}\left(\frac{e^{2x}-1}{2x} \cdot\frac{x}{\sin{x}}\cdot\frac{1}{\sin{x}}\right)=\infty$$

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  • $\begingroup$ @Salahamam_ Fatima Thank you! $\endgroup$ – Michael Rozenberg Jun 26 '17 at 17:53
  • $\begingroup$ This one make some sense, actually it is easy to understand. +1 $\endgroup$ – Vidyanshu Mishra Jun 26 '17 at 18:38
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No, it is not correct, because $(1-\cos x)'=\sin x$.

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HINT: we have $$\lim_{x\to 0}\frac{e^x-e^{-x}}{1-\cos(x)}=\infty$$

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  • $\begingroup$ The limit approaches $\infty$ from different sides on different sides of 0 (i.e. $-\infty$ on one side and $\infty$ on the other) $\endgroup$ – alphacapture Jun 26 '17 at 17:39
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There is no need to invoke De l'H$\hat{\text{o}}$pital, here rendered through

 $\hat{\text{o}}$ 

$$\frac{e^x-e^{-x}}{1-\cos(x)} = \frac{2\sinh(x)}{x}\cdot\frac{x^2}{1-\cos x}\cdot\frac{1}{x}=\frac{\sinh x}{x}\left(\frac{2x^2}{2\sin^2\frac{x}{2}}\right)\frac{1}{x} $$ and since $\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\sinh x}{x}=1$ the given limit does not exist: $$ \lim_{x\to 0^{\pm}}\frac{e^x-e^{-x}}{1-\cos(x)} = \pm\infty.$$

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  • $\begingroup$ $\lim\limits_{x\rightarrow0}\frac{1}{x}=\infty$, $\lim\limits_{x\rightarrow0^+}\frac{1}{x}=+\infty$ and $\lim\limits_{x\rightarrow0^-}\frac{1}{x}=-\infty$. I think it's better. What do you think? $\endgroup$ – Michael Rozenberg Jun 26 '17 at 17:50
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    $\begingroup$ @MichaelRozenberg: I think this is a matter of taste/conventions. I never used $\lim_{x\to 0}f(x)=\infty$, without a sign. If the limit from the right is different from the limit from the left I usually just say that the limit does not exist. $\endgroup$ – Jack D'Aurizio Jun 26 '17 at 17:54

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