$\text{⨍ }\left(\mathrm{x}\right)\text{=x cos}\left(\frac{1}{{\mathrm{x}}^2}\right)\text{, x≠0, ⨍ }\left(0\right)\text{=1,}$ at $\mathrm{x}=0 \text{⨍ }$ is

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{c}\mathrm{x}\left(0\right), \left(\frac{1}{{\mathrm{x}}^2}\right):\mathrm{x}\ne 0\\ 1\mathrm{ }:\mathrm{x}=0\end{array}\right.$

$\underset{\mathrm{x}\to 0}{\lim } \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}⟶0}{\lim } \mathrm{xcos} \left(\frac{1}{{\mathrm{x}}^2}\right)$

$=0\left(\mathrm{finite}\mathrm{ }\mathrm{value}\right)$

$=0$

$\therefore \underset{\mathrm{x}\to 0}{\lim } \mathrm{f}\left(\mathrm{x}\right)\ne \mathrm{f}\left(0\right)\Rightarrow \mathrm{f}\mathrm{ }$ is discontinuous at  x = 0

Option (2) is correct+++++++