Limit of the form $0/0$

Question

How do I find $$\lim_{x \to 0}\frac{\cos(\sin x)-\cos x}{x^4}$$ I tried rewriting $\cos$ in the form of $\sin$ and then applying $\sin x/x$ form, but it doesn't work out. Does it?

Answer 1

Hint

Taylor series could be useful. Start with $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Then use the Taylor series of $\cos(y)$; for the first term in numerator replace in the result $y$ by $x-\frac{x^3}{6}$ and for the second term in numerator replace in the result $y$ by $x$.

I am sure that you can take from here.

Answer 2

Hint : Apply L. Hospital's Rule since the limit is $\frac{0}{0}$. If the resulting form is again $\frac{0}{0}$ or $\frac{\infty}{\infty}$, continue the same process. $$\lim_{x \to 0}\dfrac{f(x)}{g(x)}=\lim_{x \to 0}\dfrac{f'(x)}{g'(x)}$$

Answer 3

First use Taylor series $$\sin x=x-\frac{x^3}{6}+o(x^3)$$ to get $$\lim_{x\to 0}\frac{x-\sin x}{x^3}=\frac{1}{6}$$ Next we can proceed as follows \begin{align} L&=\lim_{x\to 0}\frac{\cos \sin x - \cos x}{x^4}\notag\\ &=\lim_{x\to 0}\dfrac{2\sin\left(\dfrac{x+\sin x}{2}\right)\sin\left(\dfrac{x-\sin x}{2}\right)}{x^4}\notag\\ &=\lim_{x\to 0}2\cdot\dfrac{\sin\left(\dfrac{x+\sin x}{2}\right)}{\dfrac{x+\sin x}{2}}\cdot\dfrac{x+\sin x}{2x}\notag\\ &\,\,\,\,\dfrac{\sin\left(\dfrac{x-\sin x}{2}\right)}{\dfrac{x-\sin x}{2}}\cdot\dfrac{x-\sin x}{2x^3}\notag\\ &=2\cdot 1\left(\frac{1}{2}+\frac{1}{2}\right)1\cdot\frac{1}{12}=\frac{1}{6}\notag \end{align}