Limit as $x\to 0$ of a function $$\lim_{x\rightarrow 0} \frac{e^{3x}-1}{\cos \left( \sqrt{x}\right)-1}$$
Does anyone know how to solve this limit? It gives 0/0 which normally I would use l'Hopital's rule but it seems to lead me on an endless loop which I don't know how to break out of.
 A: L'Hopital works.
$\lim_{x \to 0} \frac{e^{3x}-1}{cos(\sqrt{x})-1} = \lim_{x\to 0} \frac{3e^{3x}}{-\frac
{1}{2} \frac{\sin (\sqrt{x})}{\sqrt{x}}}= \frac{3}{-\frac{1}{2}*1} = -6.
$
A: Let $\sqrt x=2h,x=4h^2$
$$\dfrac{e^{12h^2}-1}{\cos2h-1}=-\dfrac{e^{12h^2}-1}{12h^2}\left(\dfrac h{\sin h}\right)^2\dfrac{12}2$$
A: Hint:
$$\lim_{x\to0}{e^{3x}-1\over\cos(\sqrt x)-1}=\lim_{u\to0^+}{e^{3u^2}-1\over\cos u-1}$$
One or two rounds of L'Hopital now does the trick.
A: Hint:


*

*When $x$ is small, $e^x \approx 1+x$ and $\cos x \approx 1-\frac{x^2}2$.

A: Use series expansion for $e^{3x}$ and $\cos\sqrt{x}.$
$\displaystyle\lim_{x\rightarrow 0} \dfrac{e^{3x}-1}{\cos \left( \sqrt{x}\right)-1}=\displaystyle\lim_{x\rightarrow 0} \dfrac{\left(1+3x+\frac{(3x)^2}{2!}+\cdots\right)-1}{\left(1-\frac{x}{2!}+\frac{x^2}{4!}+\cdots\right)-1}=\displaystyle\lim_{x\rightarrow 0} \dfrac{3x+O(x^2)}{-\frac{x}{2}+O(x^2)}=\displaystyle\lim_{x\rightarrow 0} \dfrac{3x}{-\frac{x}{2}}=6$
A: You can solve the limit without L’Hôpital as well, such as via the series expansion:
$$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-… \implies \cos x \sim 1-\frac{x^2}{2}; \quad x\to 0$$
$$e^x = 1+x+\frac{x^2}{2!}+… \implies e^x \sim 1+x; \quad x\to 0$$
Hence, for $x \to 0$, $\cos\sqrt{x} \sim 1-\frac{x}{2}$ and $e^{3x} \sim 1+3x$. Hence, the limit simplifies to
$$\frac{1+3x-1}{1-\frac{x}{2}-1} = \frac{3x}{-\frac{x}{2}} = -6$$
