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I'm wondering how to compute the limit (if it exists) of $$\lim_{x\rightarrow 0} x^2 \csc(1/x).$$ I'm pretty sure that the limit does not exist, as the plot of $\csc(1/x)$ suggests that it is alternating as $x\rightarrow 0^+$. However, I'm not sure how to formally show this. Any hints?

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2 Answers 2

up vote 9 down vote accepted

In order for $\lim_{x\to 0+}f(x)$ to exist, there must be an interval $(0,a)$, with $a$ positive, such that $f(x)$ is defined for all $x$ in the interval $(0,a)$.

But if $x=\dfrac{1}{n\pi}$, where $n$ is a positive integer, then $\csc(1/x)$ is not defined at $x$. So there are positive $x$, arbitrarily close to $0$, at which $\csc(1/x)$ is not defined. It follows that there is no $a$ with the desired property.

Remark: We took advantage of a technical point to give an answer. But it is not hard to see that in addition, by taking $x$ close enough to $\dfrac{1}{n\pi}$, we can make $x^2\sec{(1/x)}$ arbitrarily large positive or negative.

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There’s a shortcut that you can use, thanks to the fact that there are positive real numbers arbitrarily close to $0$ at which $f$ is not defined, but I’ll show you how you could attack the problem even if such a shortcut weren’t available.

You know that $\sin x=0$ when $x$ is an integral multiple of $\pi$. Suppose that $x_n$ is just a little more than $\frac1{n\pi}$; we’ll worry about just how much a little is later. Then $|\sin 1/x_n|$ is very close to $0$, so $|\csc 1/x_n|$ is very large. And $x_n^2\ge\frac1{n^2\pi^2}$, so $|f(x_n)|\ge\frac{|\csc 1/x_n|}{n^2\pi^2}$. If we can choose $x_n$ so that $|\csc 1/x_n|\ge n^3$, we’ll be in business: we’ll have $|f(x_n)|\ge\frac{n}{\pi^2}$ for every $n$, showing that $f$ cannot have a finite limit from the right at $0$.

To get $|\csc 1/x_n|\ge n^3$, we need to get $|\sin 1/x_n|\le\frac1{n^3}$. Remember that $x_n$ is supposed to be just a little more than $\frac1{n\pi}$, so $1/x_n$ must be just a little less than $n\pi$, close enough so that $|\sin 1/x_n|\le\frac1{n^3}$. Remembering that $\sin x\approx x$ for small $x$, we try $1/x_n=n\pi-\frac1{n^3}$. Then

$$\begin{align*} |\sin 1/x_n|&=\left|\sin n\pi\cos\frac1{n^3}-\sin\frac1{n^3}\cos n\pi\right|\\ &=\left|\sin\frac1{n^3}\right|\\ &\le\frac1{n^3}\;, \end{align*}$$

exactly as we wanted. Recapitulating, we have $|\sin 1/x_n|\le\frac1{n^3}$, so $|\csc 1/x_n|\ge n^3$, and $x_n>\frac1{n\pi}$, so

$$|f(x_n)|=x_n^2|\csc 1/x_n|>\frac{n^3}{n^2\pi^2}=\frac{n}{\pi}\;.$$

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