# How to understand that the $\lim_{x\to 0}\frac1x\cos(\frac1x)$ does not exist.

I'm studying the $\lim_{x\to 0}\frac1x\cos(\frac1x)$. Can I understand that the limit does not exist simply splitting it into two parts? That is, $$\lim_{x\to 0}\frac1x\cos(\frac1x)=\lim_{x\to 0}\frac {\cos(\frac1x)}{x}.$$ Now, the original limit does not exist because the $\lim_{x\to 0}\cos(\frac1x)$ does not exist. I am using a quotient rule.

Another example could be the $\lim_{n\to\infty}\sqrt n\cos(nx)$ because I can split it in the following way $$\lim_{n\to\infty}\frac{\cos(nx)}{1/\sqrt n}.$$

Question. Is my little argument correct? Should I use something else (i.e another technique) to study this limit in order to understand if it does exist or not?

Thank you.

## 4 Answers

I fail to see an argument in your question and I would strongly urge against writing anything like $$\lim_{x\to0}g(x)$$ when one is not assured the limit exists (even worse, when one tries to show it does not exist).

Here $g(x)=\frac1x\cos\left(\frac1x\right)$ hence a simple way to show the limit when $x\to0$ does not exist is to compute $$g\left(\frac1{2n\pi}\right)=2n\pi,\qquad g\left(\frac1{2n\pi+\pi/2}\right)=0.$$ Since the first sequence converges to $+\infty$ and the second sequence converges to $0$, and since these are different, the function $g$ has no limit at $0$ (actually, this even shows the stronger statement that $g$ has no limit at $0^+$).

True, this assumes one somehow can guess two adapted sequences but then anyway, at one point or another, one has to see what happens to solve the exercise.

• The sequences $\{\frac{1}{2n\pi}\}$ and $\{\frac{1}{2n\pi+\pi/2}\}$ both converge to $0$. The limit does not exist because if the limit existed, the function of two such sequences would be the same. Is my understanding correct? Commented Jan 14, 2014 at 9:51
• "if the limit OF $g$ AT $0$ existed, the LIMITS of THE IMAGES BY $g$ OF two such sequences would be the same." Exactly.
– Did
Commented Jan 14, 2014 at 9:54
• Does the simple use of the symbol $\lim_{x\to c} f(x)$ imply that I am thinking that the limit exist? Commented Jan 14, 2014 at 9:59
• Yes, you are even asserting it does exist.
– Did
Commented Jan 14, 2014 at 9:59

if you choose a sequence $x_n=\frac{1}{2n \pi}$ then you get $2n\pi \cos(2n\pi)=2n\pi \to +\infty$

if you choose a sequence $x_n=\frac{1}{(2n+1) \pi}$ then you get $(2n+1)\pi \cos((2n+1)\pi)=-(2n+1)\pi \to -\infty$

Set $t:=1/x$, so your limit becomes:

$$\lim_{t\to\infty} t \cos{t}.$$

This limit does not exist, does it?

I hope somebody can help you more.

Cheers.

Your argument is flawed: $\displaystyle\lim_{x\to+\infty}{\cos(x)\over x} = 0$. But suppose that your limit exists and consider (using product rule) $\displaystyle\lim_{x\to 0}x{1\over x}{\cos({1\over x})}$.