Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + 2x \cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right) $$  The derivative of the denominator is $\cos(x)$. So,  $$\lim\limits_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)} = \lim\limits_{x\rightarrow 0}\displaystyle\frac{\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)}{\cos(x)}$$
Is that right so far?
Thanks for the help in advance.
Best Regards,
Ahmed Hossam

As we know that $$\lim_{x\to 0}{x\over \sin x}=1$$therefore $$\lim_{x\to 0}{x^2\cos {1\over x}\over \sin x}=\lim_{x\to 0}x\cos{1\over x}=\lim_{u\to \infty}{\cos u\over u}=0$$
