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?
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2 Answers
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You can't squeeze it but you can estimate the given term by something divergent as follows for $x<0$:
$$\frac{\sin\left(-\frac{3}{x}\right)-4}{x}\stackrel{x<0\Rightarrow|x|=-x}{=}\frac{4+\sin\left(\frac{3}{x}\right)}{|x|}\geq \frac 3{|x|}\stackrel{x\to 0^-}{\longrightarrow}+\infty$$
answered Feb 6, 2020 at 2:19
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$\begingroup$ is there any reason why we cannot use squeeze therom? $\endgroup$ Feb 6, 2020 at 2:21
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$\begingroup$ @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 $+\infty$. $\endgroup$ Feb 6, 2020 at 2:23
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$\begingroup$ @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 $\endgroup$ Feb 6, 2020 at 2:26
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$\begingroup$ thanks sir last question how you evaluate 4+sin(3/𝑥)=3 ? this is the actual thing I want to learn $\endgroup$ Feb 6, 2020 at 2:27
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$\begingroup$ 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. $\endgroup$ Feb 6, 2020 at 2:28
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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 $x \rightarrow 0$.
