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How would I show that the following limit does not exist?
$$\lim_{ x \to 0 }2\sin\left(\frac{1}{x^2}\right)$$
Do I just state that $\frac{1}{0^2}$ is undefined, therefore the limit doesn't exist? Or maybe I have to use the fact that sin is periodic? I'm new to limits sorry for my limited knowledge.
Let $x_n=(n \pi)^{-1/2}$. Then $x_n \to 0$ and $2 \sin(\frac{1}{x_n^2})=0 $ for all $n$.
It is your turn to find a sequence $(z_n)$ with $z_n \to 0$ and $2 \sin(\frac{1}{z_n^2})=2 $ for all $n$.
These two sequences show that $\lim_{ x \to 0 }2 \sin(\frac{1}{x^2})$ does not exist.
Consider the sequence $(x_n)_{n\in\mathbb N}$ defined by$$(\forall n\in\mathbb{N}):x_n=\frac1{\sqrt{n\pi}}.$$Then $\lim_{n\to\infty}2\sin(x_n)=0$. Can you provide a sequence $(y_n)_{n\in\mathbb N}$ such that $\lim_{n\to\infty}y_n=0$ and that $\lim_{n\to\infty}2\sin(y_n)$ exists but is different from $0$?
