Question

show that ${lim}_{x\to 0}sin\frac{1}{x}$does not exist.

Answer 1

${lim}_{x\to 0}sin\frac{1}{x}={lim}_{h\to 0}sin\frac{1}{0-h}=-{lim}_{x\to 0}sin\frac{1}{h}$

=-(An oscillating number which oscilates between -1 and 1).

So, ${lim}_{x\to 0^{-}}sin\frac{1}{x}$ does not exist.

Similarly, ${lim}_{x\to 0^{+}}sin\frac{1}{x}$ does not exist.

${lim}_{x\to 0}sin\frac{1}{x}$ does not exist.