Evaluating $\lim_{x\rightarrow 0}\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}$ How do I evaluate the limit $\lim_{x\rightarrow 0}\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}$? The only method I know to deal with these kinds of limits is L'hopital, but it doesn't seem to help here at all..
 A: The method of Taylor expansions is very elegant, but you can do this in a different way.
You surely know how to compute
$$
\lim_{t\to0}\frac{1-\cos t}{t^2}=\frac{1}{2},
\quad
\lim_{t\to0}\frac{e^t-1}{t}=1,
\quad
\lim_{t\to0}\frac{\sin t-t}{t^3}=-\frac{1}{6}
$$
From the first you get
$$
\frac{1}{2}=\lim_{x\to0}\frac{1-\cos(\sin^5x)}{\sin^{10}x}=
\lim_{x\to0}\frac{1-\cos(\sin^5x)}{x^{10}}\frac{x^{10}}{\sin^{10}x}
$$
and so
$$
\lim_{x\to0}\frac{1-\cos(\sin^5x)}{x^{10}}=\frac{1}{2}
$$
From the second you get
$$
\lim_{x\to0}\frac{e^{x^4}-1}{x^4}=1
$$
and from the third that
$$
\lim_{x\to0}\frac{\sin(x^2)-x^2}{x^6}=-\frac{1}{6}
$$
Now rewrite your limit as
$$
\lim_{x\to 0}\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}=
\lim_{x\to0}
  \frac{1-\cos(\sin^5x)}{x^{10}}
  \frac{x^4}{e^{x^4}-1}
  \frac{x^6}{\sin(x^2)-x^2}
=\frac{1}{2}\cdot 1\cdot(-6)=-3
$$
A: Hint. You have, using standard Taylor expansions, as $x \to 0$,
$$
1-\cos (\sin ^5(x))=\frac{x^{10}}2+o(x^{10})
$$
and$$
(e^{x^4}-1)(\sin(x^2)-x^2)=x^{4} \times\frac{-x^{6}}6+o(x^{10})
$$ giving

$$
\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}=-3+o(x^{10})
$$

A: $$\sin(x)=x-x^3/6+O(x^5)$$
$$\cos(x)=1-x^2/2 +O(x^4)$$
$$e^x=1+x+O(x^2)$$
Hence
$$\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}$$
is asymptotically equivalent to
$$\frac{\frac{1}{2}\sin ^{10}(x)}{(x^4)(-x^6/6)}$$
and so the limit is -3.
