# Math Help - Does the limit exist?

1. ## Does the limit exist?

$f(x) = cos(\frac{1}{x})(sin(x))$ could any1 tell me how to find the limit as $x \to 0$ i have tried every mthod i now and its getting me nowere.
Does the limit even exist?

I was thinking that as $x \to 0$, $sin(x) \to 0$ and $cos(\frac{1}{x})$ doesn't approach a limit i think it would just oscilate. So no limit would exist.

2. Originally Posted by Lipticboven
$f(x) = cos(\frac{1}{x})(sin(x))$ could any1 tell me how to find the limit as $x \to 0$ i have tried every mthod i now and its getting me nowere.
Does the limit even exist?

I was thinking that as $x \to 0$, $sin(x) \to 0$ and $cos(\frac{1}{x})$ doesn't approach a limit i think it would just oscilate. So no limit would exist.
This is what I would do:

$\lim_{x\to 0}\cos\left(\frac{1}{x}\right)\sin x=\lim_{x\to 0}{\color{red}\frac{x}{x}}\cdot\cos\left(\frac{1}{ x}\right)\sin x=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\frac{\sin x}{x}$

Now, recall that $\lim_{x\to c}f(x)g(x)=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$

So we have $\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\frac{\sin x}{x}=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot\lim_{x\to 0}\frac{\sin x}{x}$

We should familiarize ourselves with two special limits:

$\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to \infty}\frac{\cos x}{x}=0$

We need to incorporate these somehow...

We can first start off by simplifying this:

$\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot\underbrace{\l im_{x\to 0}\frac{\sin x}{x}}_1=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot1=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)$

Now if you make a substitution, say $z=\frac{1}{x}\implies x=\frac{1}{z}$, we see that as $x\to 0$, $z\to\infty$

Thus, $\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)=\lim_{z\to\infty}\ frac{\cos z}{z}=0$

Thus, $\lim_{x\to 0}\cos\left(\frac{1}{x}\right)\sin x=\color{red}\boxed{0}$

Does this make sense?

--Chris

3. I would never have thought of that
thanks

4. Originally Posted by Chris L T521
This is what I would do:

$\lim_{x\to 0}\cos\left(\frac{1}{x}\right)\sin x=\lim_{x\to 0}{\color{red}\frac{x}{x}}\cdot\cos\left(\frac{1}{ x}\right)\sin x=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\frac{\sin x}{x}$

Now, recall that $\lim_{x\to c}f(x)g(x)=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$

So we have $\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\frac{\sin x}{x}=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot\lim_{x\to 0}\frac{\sin x}{x}$

We should familiarize ourselves with two special limits:

$\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to \infty}\frac{\cos x}{x}=0$

We need to incorporate these somehow...

We can first start off by simplifying this:

$\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot\underbrace{\l im_{x\to 0}\frac{\sin x}{x}}_1=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot1=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)$

Now if you make a substitution, say $z=\frac{1}{x}\implies x=\frac{1}{z}$, we see that as $x\to 0$, $z\to\infty$

Thus, $\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)=\lim_{z\to\infty}\ frac{\cos z}{z}=0$

Thus, $\lim_{x\to 0}\cos\left(\frac{1}{x}\right)\sin x=\color{red}\boxed{0}$

Does this make sense?

--Chris

It's overkill I guess.

We know that $\lim_{x\to 0}\sin x = 0$.

We also know that cos() is always between -1 and 1:
$-1\leq \cos \frac{1}{x}\leq 1
$

If an expression has limit 0, then we may multiply it with things that are maximum 1 in absolute value, the result will still have limit 0.

Why?
Say we want to get a neighborhood of 0 where f(x) is always $\epsilon$ close or even closer to 0.
We can choose such a neighborhood for sin(x), because sin(x) has limit 0 at x=0, and the definition of the limit states just what we want.
But in f(x), we also have that cos(1/x) thing there. Okay, but we saw it is between -1 and +1, so multiplying by it can only take us closer to 0, or leave us as close as before. (It might get us on the other side, if it is negative, but the distance from 0 can not be larger than before).
So? We are done. We got the neighborhood of 0, where f(x) is always maximum $\epsilon$ far from 0. We didn't use how small $\epsilon$ is, so by definition of the limit, f(x) has the limit 0.