Calculate $\lim_{x\to0}[x^3 \sin \frac 1{x^2}]$?

Question

How do I calculate $$\lim_{x\to0}[x^3 \sin \frac 1{x^2}]$$ I'm thinking that since $$\frac{\sin x}{x} \to 1$$ Maybe I can do this as well? $$\frac{\sin \frac 1{x^2}}{\frac 1{x^2}}=x^2\sin \frac 1{x^2} \to 1$$ Which would leave me with $$\lim_{x\to0} x \cdot 1 = 0$$ Would this make sense?

Answer 1

First notice that the function $\sin(\frac{1}{x^2}) $ is bounded, i.e., $$ \bigg|\sin(\frac{1}{x^2})\big| \leq 1.$$ Hence $$ |x^3 \sin(\frac{1}{x^2})| \leq |x^3| $$ and $$ \lim_{x \rightarrow 0} |x^3| = 0.$$ So we have, by the squeeze theorem,

$$\lim_{x \rightarrow 0} x^3 \sin(\frac{1}{x^2}) = 0$$

Answer 2

The rule only applies if $x\to 0$, so it is not applicable in your case.

To calculate your limit, an even simpler thing is needed:

Hint:

For all values of $y$, $|\sin(y)|<1$.

Answer 3

Use the squeeze theorem. The function $\sin(1/x^2)$ is bounded between $-1$ and $1$ while $x^3$ tends to $0$ as $x$ tends to $0$.