Evaluate $\lim_{x\to0}(\cosh 3x)^{\csc^2x}$ I came across a problem about evaluating the following limit:
$$\lim_{x\to0}(\cosh 3x)^{\csc^2x}$$
My attempt:
I think this is about applying L'Hopital rule. Here's what I've tried:
$
\lim_{x\to0}(\cosh 3x)^{\csc^2x}\\=\lim_{x\to0}(\frac{e^{3x}+e^{-3x}}{2})^{\csc^2 x}\\=\lim_{x\to0}\frac{(e^{3x}+e^{-3x})^{\csc^2 x}}{2^{\csc^2 x}}$
Should I just apply L'Hopital Here? I am not sure because the derivative of the numerator looks really monstrous and complicated. Is there any better approach of this problem or any way to simplify? Any hint or suggestion is welcomed. Thanks a lot.
 A: You should take logarithm first. So consider
$$
g(x)=\csc^2x\ln(\cosh 3x).
$$
Expanding in powers of $x$ around the origin
$$
g(x)=\frac{\ln (1+9x^2/2+\dots)}{x^2+\dots}=\frac{9x^2/2+\dots}{x^2}
$$
Therefore, $\lim_{x\to 0}g(x)=9/2$ and your limit is $e^{9/2}$.
You can also apply L'Hopital's rule to $g$, keeping the logarithm in the numerator.
A: By standard limit we have
$$(\cosh 3x)^{\csc^2x}=\left[(1+(\cosh 3x-1))^{\frac1{\cosh 3x-1}}\right]^{\csc^2x(\cosh 3x-1)} \to e^\frac92$$
indeed since $t=\cosh 3x-1 \to 0$
$$(1+(\cosh 3x-1))^{\frac1{\cosh 3x-1}}=(1+t)^\frac1t \to e$$
and
$$\csc^2x(\cosh 3x-1)=9\frac{x^2}{\sin^2 x}\frac{\cosh 3x-1}{9x^2} \to 9\cdot 1\cdot\frac12=\frac92$$
A: Hint: Take logarithm. You have to find $\lim \frac {\ln (\cosh (3x)} {\sin^{2}(x)}$. This has $\frac 0 0$ form. Apply L'Hopital's Rule to get $\lim \frac {3\sinh (3x)} {(2\sin x \cos x) (\cosh (3x))}$. At this stage remove $\cosh (3x)$ and $\cos x$ because they tend to $1$. Now you just need another simple application of L'Hopital's Rule.  You should get $e^{\frac 9 2}$ as the limit.
