(Verification)$\lim\limits_{x\to 0}{e^x-e^{-x} \over 1- \cos x} =2$

Question

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

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?

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

Answer 1

$$\lim\limits_{x\to 0}{e^x-e^{-x} \over 1- \cos x}=\lim\limits_{x\to 0}\frac{(e^{2x}-1)(1+\cos{x})} {e^x(1- \cos^2 x)}=\lim\limits_{x\to 0}\frac{2(e^{2x}-1)} {\sin^2x}=$$ $$=4\lim\limits_{x\to 0}\left(\frac{e^{2x}-1}{2x} \cdot\frac{x}{\sin{x}}\cdot\frac{1}{\sin{x}}\right)=\infty$$

Answer 2

No, it is not correct, because $(1-\cos x)'=\sin x$.

Answer 3

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

Answer 4

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: $$ \lim_{x\to 0^{\pm}}\frac{e^x-e^{-x}}{1-\cos(x)} = \pm\infty.$$