Limit without L'Hôpital's theorem or squeeze theorem

Question

I was trying to solve: $$\lim _{x\to 0^-}\left(\frac{\sin\frac{-3}{x}-4}{x}\right)$$ from the squeeze theorem we know that $$3\le \sin\frac{-3}{x}\le -3$$ My question since limit is approaching to left side of $0$. Should I apply limit on both side of like $$\lim _{x\to 0^-}-3\le \sin\frac{-3}{x}\le \lim _{x\to 0^-}3$$ or one side $$\lim _{x\to 0^-}-3$$ if there any other easy evaluation possible?

Answer 1

You can't squeeze it but you can estimate the given term by something divergent as follows for $x<0$:

$$\frac{\sin\left(-\frac{3}{x}\right)-4}{x}\stackrel{x<0\Rightarrow|x|=-x}{=}\frac{4+\sin\left(\frac{3}{x}\right)}{|x|}\geq \frac 3{|x|}\stackrel{x\to 0^-}{\longrightarrow}+\infty$$

Answer 2

Actually, you can bound $\sin(\frac{-3}{4})$ by $-1$ and $1$, ie, $-1\leq \sin(\frac{-3}{4})\leq 1$. So your function, say $f(x)$ is bounded by $g(x)$ and $h(x)$, where $g(x)=\frac{-5}{x}$ and $h(x)=\frac{-3}{x}$, ie, $$g(x)\leq f(x) \leq h(x)$$

But, the limit of both $g(x)$ and $h(x)$ is $\infty$ as $x \rightarrow 0$, so $f(x)$ must also have the limit of $\infty$ as $x \rightarrow 0$.