Define function g on [-1, 1] by $g(x) = $cos($1/x$), if $\,x\neq 0\,$ and g(x) = 0, if x = 0.

Question

Define function $g$ on $[-1, 1]$ by $g(x) = \cos(1/x)$, if $x \neq 0$ and $g(x) = 0$, if $x = 0$.

Prove that g is not continuous 0.

I have looked at using the negation of the definition:

For $\{x_n\}_{n=1}^\infty \subset I$ with $\lim_{x\to c} x_n=c$ then $\lim_{x\rightarrow c} f(x)=f(c)$

Answer 1

Let $x_n=\frac{1}{n\pi}$. Then $x_n\to 0$ but $g(x_n)$ does not converge.

Answer 2

Let Xn= 1/2πn, n∈N. Then Xn→0 and g(Xn)→1. We have a sequence {Xn} in R that converges to 0 but lim g(Xn)≠ g(0). So g is not continuous at 0.