Solve $\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$ Find the limit: $$\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$$
After treating it with l'Hopital rule, we get:
$$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}-\cos\frac{1}{x}}{\cos x}}$$
Now, the numerator of fraction doesn't have a limit, so I can't use l'Hopital rule again.
What to do?
I can split it into two fractions, but I'm not sure how would it help:
$$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}}{\cos x}} - \lim_{x\to0}{\frac{\cos\frac{1}{x}}{\cos x}}$$
 A: $$0\leq \left| \frac{x^2 \text{sin}\frac{1}{x}}{\text{sin} x} \right| \leq \left| \frac{x^2}{\sin x} \right|$$
You can now apply L'Hôpital's rule to $\frac{x^2}{\text{sin}x}$ to show the limit of this as $x \rightarrow 0$ is $0$, then use the squeeze rule.
A: You can only split if both limits exist. In this case, it's fine. However, l'Hospital doesn't help here. The derivative of $\sin \frac{1}{x}$ wildly diverges so it's useless. What you need to realize is, that $x\sin\frac{1}{x}$ is a product of a bounded function and a function with limit $0$, which tells you about the limit.
The part $\frac{x}{\sin x}$ converges to 1, that's known.
A: Rewrite the function as
$$ \frac{\left(\frac{\sin \frac1x}{\frac1x}\right)  }{\left(\frac{\sin x}{x}\right)} $$
Now think about the behavior of the simpler function $g(x)=\sin x/ x$ both as $x\rightarrow 0$ and as $x\rightarrow \pm \infty$.
As $x\rightarrow 0$, $g(x) \rightarrow 1$.
As $x\rightarrow \pm\infty$, $g(x)\rightarrow 0$.
These are precisely the behaviors of the denominator and numerator, respectively, as $x\rightarrow 0$. So the limit of the original expression as $x\rightarrow 0$ is $0/1=0$.
A: $$ \lim_{x\to 0} \frac{x^2\sin\left(\frac{1}{x}\right)}{\sin(x)} = \left[\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)\right]\left[\lim_{x\to 0}\frac{x}{\sin(x)}\right]= \lim_{x\to 0} x\sin\left(\frac{1}{x}\right) $$
Let $z=\frac{1}{x}$. Since $x\to 0$, then $z\to\infty$. So now we have
$$ \lim_{z\to \infty} \frac{\sin\left(z\right)}{z} =0 $$
