Trigonometrical limit $$\lim\limits_{x \to 0} \frac{x^4\cos(\frac{1}{x})} {\sin^3x},$$
I have a problem with this limit. I tried to use L'Hôpital's rule but it is not effective. Please help.
 A: Hint: $\lim_{x\to 0}\dfrac{x\cos(\frac{1}{x})}{\sin^3(x)x^{-3}}$
Now, find the limit of the numerator via the sandwich theorem and the limit of denominator is a well known result : $\lim_{x\to 0}\frac{x}{\sin(x)}=1$.
Hope this helps!
A: $$\lim\limits_{x → 0} \frac{x^4 \cos \left(x^{-1}\right)}{\sin ^3\left(x\right)} \times \frac{x^{-3}}{x^{-3}}$$

$$\lim\limits_{x → 0} \frac{x\cos \left(x^-1\right)}{x^{-3}\sin ^3\left(x\right)}$$

$$\lim\limits_{x → 0} \frac{ \frac{\cos \left(x^{-1}\right)}{x^{-1}} }{ \frac{\sin ^3\left(x\right)}{x^3} } $$


*

*Using $$\lim\limits_{x → 0} \frac{\sin \left(x\right)} {x}=1 \quad \textrm{ & } \lim\limits_{x → 0} \frac{\cos \left(1/x\right)} {1/x}=0$$

$$\therefore \frac{ \lim\limits_{x → 0} 
\frac{\cos \left(x^{-1}\right)}{x^{-1}} }{ \lim\limits_{x → 0} \left( \frac{\sin \left(x\right)}{x} \right)^3 } =0$$
A: The limit is zero, even in the more general case where $n\geq1$ :
$$\lim_{x\rightarrow0}\left|\frac{x^{n}\cos\left(\frac{1}{x}\right)}{\sin^{n-1}\left(x\right)}\right|=\lim_{x\rightarrow0}\left|\frac{x^{n-1}}{\sin^{n-1}\left(x\right)}\right|\left|x\right|\left|\cos\left(\frac{1}{x}\right)\right|=\lim_{x\rightarrow0}\left|\frac{1}{\mathrm{sinc}\left(x\right)}\right|^{n-1}\lim_{x\rightarrow0}\left|x\right|\left|\cos\left(\frac{1}{x}\right)\right|
 $$
$$\leq\left(\lim_{x\rightarrow0}\left|\frac{1}{\mathrm{sinc}\left(x\right)}\right|\right)^{n-1}\lim_{x\rightarrow0}\left|x\right|.1=1^{n-1}.0=0$$
where $\mathrm{sinc}\left(x\right)=\frac{\sin\left(x\right)}{x}$ denotes the cardinal sine. At the limit $x\rightarrow0$, one has
$$\lim_{x\rightarrow0}\frac{\sin\left(x\right)}{x}=\lim_{x\rightarrow0}\frac{\sin\left(x\right)-\sin\left(0\right)}{x-0}=\cos\left(0\right)=1.
 $$
Of course, one can use the continuity of the composition $x\mapsto\left(\frac{1}{\mathrm{sinc}\left(x\right)}\right)^{n-1}
 $ near $x=0$ to take the limit.
