A difficulty in using squeeze theorem.

Question

Given that: $\forall x \in \mathbb{R}, \sin^2(3x) \leq x^2f(x)\leq 3x^2 + 6\sin(x^2),$ then $$ \begin{equation*} \lim_{x \rightarrow 0} f(x) = \end{equation*}$$

i)0

ii)9

iii)$\infty$

iv) The limit does not exist.

My idea is to divide by $x^2$ the whole inequality and so the answer is 9, but I am hesitated to do this because I am not sure that $x^2 \ne 0$.

Answer 1

Limit $x\to 0$ tells you, how your function behaves when $x$ approaches $0$, but is not $0$, so your approach is correct.

For example, if

$h(x)=\begin{cases} 1& x\neq 0\\ 2& x = 0 \end{cases}$

then $\lim_{x\to 0} h(x) = 1$, even though $h(0) = 2$.

Answer 2

Yes, that's the way to do it. In a limit $x\to0$, you are assuming $x\ne0$, so your idea works fine.