Calculating limits for indeterminate forms I am asked to calculate the limit as $x\to0$ of:
$$
\frac{e^x+e^{-x}-2}{1-\cos(3x)}
$$
I believe this is an "infinity/infinity" problem where i could directly apply L'Hopital's rule. Is this right? how would this limit be found?
 A: The numerator and denominator both approach $0$.  Application of L'Hopital's rule to this problem should be routine, but you'll need to do it twice.
A: 1). L'Hopital's rule is a hammer. IMO, it should not be a reflex, definitely not the first thing to try and apply.
2). Do you know Taylor series? If you do, recall that
$e^x = 1+x+\frac{x^2}{2}+o(x^2)$, $e^{-x} = 1-x+\frac{x^2}{2}+o(x^2)$, and $\cos 3x = 1- \frac{9x^2}{2} + o(x^2)$. This gives
$$
\frac{1+x+1-x-2 + x^2 + o(x^2)}{1- 1 + \frac{9x^2}{2} + o(x^2) }  = \frac{x^2 + o(x^2)}{\frac{9x^2}{2} + o(x^2) } = \frac{1 + o(1)}{\frac{9}{2} + o(1) } \xrightarrow[x\to 0]{} \frac{2}{9}
$$
for your expression.
A: You can use L'Hopital's but you have to use it twice. First we see
$$L=\lim_{x\to0}\frac{e^x+e^{-x}-2}{1-\cos3x}=\lim_{x\to0}\frac{e^x-e^{-x}}{3\sin3x}$$
and then once more (as this is still $0/0$):
$$L = \lim_{x\to0}\frac{e^x+e^{-x}}{9\cos3x}=\frac{2}{9}$$
by direct substitution. 
A: $$\frac{e^x+e^{-x}-2}{1-\cos3x}=\left(\frac{e^x-1}x\right)^2\cdot\frac1{e^x}\cdot\frac1{\dfrac{1-\cos3x}{x^2}}$$
Now $\lim_{x\to0}\dfrac{e^x-1}x=1$
and for the last part, use Find $\lim_{x \to 0} \frac{1-\cos5x}{1-\cos3x}$
