# Limit without L'Hôpital's theorem or squeeze theorem

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?

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

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

• is there any reason why we cannot use squeeze therom? Feb 6, 2020 at 2:21
• @FeemoFellow : Squeezing is used to show that an expression has a FINITE limit. But here you have an expression which is diverging to $+\infty$ for $x\to 0^-$. So, you cannot squeeze it because it is "running away" to $+\infty$. Feb 6, 2020 at 2:23
• @FeemoFellow : Maybe the graph can convince you that a proper squeezing is not possible. wolframalpha.com/input/?i=plot+%28sin%28-3%2Fx%29-4%29%2Fx Feb 6, 2020 at 2:26
• thanks sir last question how you evaluate 4+sin(3/𝑥)=3 ? this is the actual thing I want to learn Feb 6, 2020 at 2:27
• But what you can do is the following (but this is hardly "squeezing") in the sense of the word: $$\frac 5{|x|}\geq\frac{4+\sin\left(\frac{3}{x}\right)}{|x|}\geq \frac 3{|x|}$$ But, since the lower bound goes to $+\infty$, you do not need the upper bound. Feb 6, 2020 at 2:28

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 $$x \rightarrow 0$$.

• a good approach also thank sir Feb 6, 2020 at 2:35