Evaluating $\lim_{x\rightarrow 0}\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}$

Question

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..

Answer 1

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

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

Answer 2

$$\sin(x)=x-x^3/6+O(x^5)$$

$$\cos(x)=1-x^2/2 +O(x^4)$$

$$e^x=1+x+O(x^2)$$

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.

Answer 3

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 $$ \lim_{x\to 0}\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}= \lim_{x\to0} \frac{1-\cos(\sin^5x)}{x^{10}} \frac{x^4}{e^{x^4}-1} \frac{x^6}{\sin(x^2)-x^2} =\frac{1}{2}\cdot 1\cdot(-6)=-3 $$