Prove that lim x -> 0 sin (1/x) doesn't exist

[Portuga]

Homework Statement

prove that
$$<br /> \lim_{x\rightarrow 0} \sin \left( \frac{1}{x} \right)<br />$$ doesn't exist.

Homework Equations

$$\lim_{x\rightarrow0}\sin\left(\frac{1}{x}\right)=\lim_{u\rightarrow\infty}\sin u$$

The Attempt at a Solution

My strategy to solve this problem is to make $u \rightarrow \infty$ through different paths and show that the limits are different.
So, first one is $a_{u}=\frac{\pi}{2}+2\pi u,\,u\in\mathbb{N}$. Second one is $b_{u}=u\pi,\,u\in\mathbb{N}$.
Both these have the same behavior as $u \rightarrow \infty$:
$$<br /> \lim_{u\rightarrow\infty}a_{u}=\lim_{u\rightarrow\infty}b_{u}=\infty.<br />$$
For with first one,
$$\begin{aligned}\lim_{u\rightarrow\infty}\sin a_{u} & =\lim_{u\rightarrow\infty}\sin\left(\frac{\pi}{2}+2\pi u\right)\\ & =\lim_{u\rightarrow\infty}\sin\left(\frac{\pi}{2}\right)\cos2\pi u+\sin2\pi u\cos\left(\frac{\pi}{2}\right)\\ & =\lim_{u\rightarrow\infty}\sin\left(\frac{\pi}{2}\right)\cos2\pi u+\lim_{u\rightarrow\infty}\sin2\pi u\cos\left(\frac{\pi}{2}\right)\\ & =\lim_{u\rightarrow\infty}\cos2\pi u+\lim_{u\rightarrow\infty}\sin2\pi u\cos\left(\frac{\pi}{2}\right). \end{aligned}$$

The limit $\lim_{u\rightarrow\infty}\sin2\pi u\cos\left(\frac{\pi}{2}\right)$ tends to 0, as $\cos\left(\frac{\pi}{2}\right)$ tends to 0, and $\sin2\pi u$ is limited both upper and lower bound.

Therefore, $$\lim_{u\rightarrow\infty}\sin a_{u}=1.$$

Now, $$\lim_{u\rightarrow\infty}\sin b_{u}=\lim_{u\rightarrow\infty}\sin u\pi=0,$$ as, $\forall u, u\pi$ is an multiple of $\pi$, which $\sin$ is null.
Then, $$\lim_{u\rightarrow\infty}\sin b_{u}=0.$$
Because of these two different results for different paths, it's true that $$\nexists\lim_{x\rightarrow0}\sin\frac{1}{x}.$$

Am I correct?
Thanks in advance!

 

[PeroK]

It looks okay, but could have been a lot simpler:

$\sin(\frac{\pi}{2} + 2\pi u) = \sin{\frac{\pi}{2}} = 1$

Also, you could just let $a_u = \frac{1}{\frac{\pi}{2} + 2\pi u}$ and keep the limits going to $0$.

 

[mfb]

You are overthinking this. The individual functions are constant, their limit is easy to find.
Directly using sequences for the original function is better, because it avoids issues with the sign change at 0.

 

[Portuga]

You are overthinking this. The individual functions are constant, their limit is easy to find.
Directly using sequences for the original function is better, because it avoids issues with the sign change at 0.

You mean, I should use $a_{u}=\frac{1}{\frac{\pi}{2}+2\pi u},\,u\in\mathbb{N}$ and $\lim_{x\rightarrow0}\sin\frac{1}{x}$ since the beginning, correct?

 

[Portuga]

It looks okay, but could have been a lot simpler:

$\sin(\frac{\pi}{2} + 2\pi u) = \sin{\frac{\pi}{2}} = 1$

Also, you could just let $a_u = \frac{1}{\frac{\pi}{2} + 2\pi u}$ and keep the limits going to $0$.

 

[PeroK]

You mean, I should use $a_{u}=\frac{1}{\frac{\pi}{2}+2\pi u},\,u\in\mathbb{N}$ and $\lim_{x\rightarrow0}\sin\frac{1}{x}$ since the beginning, correct?

I thought you were using the idea of finding two sequences that converge to 0 but for which the two sequences of function values converge to different values. That looks fine as an approach. Changing the limit to $\infty$ requires a bit more justification.

 

[Portuga]

I thought you were using the idea of finding two sequences that converge to 0 but for which the two sequences of function values converge to different values. That looks fine as an approach. Changing the limit to $\infty$ requires a bit more justification.

Yes, that's exactly what's my initial approach was: two sequences converging to same value, but producing different limits with $\sin \left( \frac{1}{x} \right)$.

 

[PeroK]

Homework Equations

$$\lim_{x\rightarrow0}\sin\left(\frac{1}{x}\right)=\lim_{u\rightarrow\infty}\sin u$$

One minor problem is that:

$$\lim_{x\rightarrow0+}\sin\left(\frac{1}{x}\right)=\lim_{u\rightarrow\infty}\sin u$$

This is not fatal to your proof, but what you have is not quite right. It's better to stick with sequences going to 0.