Proving limits with the epilson delta definition

Question

1. I want to prove that $\lim\limits_{x\to 0} \sin\left(\frac{1}{x}\right)$ does not exist using the $\epsilon-\delta$ definition of limits. I know I need to algebraically prove that for $0<|x|<\delta$, $|\sin(1/x) - L| ≥ \epsilon$ , but I'm not sure how.

2. How can you show that the $\lim\limits_{x\to 0} x\sin \left(\frac{1}{x}\right) = 0$ using the $\epsilon-\delta$ definition of limits?

Answer 1

Here is a definition of limit.

Both problems are easy: I am giving a hint for $2nd$ problem. $$|f(x)-f(0)|=|xsin\frac{1}{x}|\leq |x-0|<\delta$$, Choose $\delta=\epsilon$ $$|f(x)-f(0)|\leq \epsilon$$

Answer 2

For first question is enough to consider sequences $x_n=\frac{1}{2 \pi n}$ and $y_n=\frac{1}{2 \pi n+\frac{\pi}{2}}$. Both tends to $0$ but $\sin$ have different fixed values $0,1$ in them, so it is not possible to have limit.

For second, as already mentioned, is enough to use $|\sin x| \leqslant 1$ and use $\lim\limits_{x\to 0} x = 0$.