Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.