# The value of

Question:

$f(x)=\left\{\begin{array}{ll}|x| \cos \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.$ at $\mathrm{x}=0$

Solution:

Checking the right hand and left hand limits for the given function, we have

$\lim _{x \rightarrow 0^{-}} f(x)=|x| \cos \frac{1}{x}$

$=\lim _{h \rightarrow 0}|0-h| \cos \frac{1}{(0-h)}=\lim _{h \rightarrow 0} h \cos \frac{1}{h}$

$=0 \quad\left[\because \cos \frac{1}{x}\right.$ oscillate between $-1$ and 1$]$

$\lim _{x \rightarrow 0^{+}} f(x)=|x| \cos \frac{1}{x}$

$=\lim _{h \rightarrow 0}|0+h| \cos \frac{1}{(0+h)}=\lim _{h \rightarrow 0} h \cdot \cos \frac{1}{h}=0$

$\lim _{x \rightarrow 0} f(x)=0$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} f(x)=0$

$\lim _{x \rightarrow 0^{-}} f(x)=|x| \cos \frac{1}{x}$

$=\lim _{h \rightarrow 0}|0-h| \cos \frac{1}{(0-h)}=\lim _{h \rightarrow 0} h \cos \frac{1}{h}$

$=0 \quad\left[\because \cos \frac{1}{x}\right.$ oscillate between $-1$ and 1$]$

$\lim _{x \rightarrow 0^{+}} f(x)=|x| \cos \frac{1}{x}$

$=\lim _{h \rightarrow 0}|0+h| \cos \frac{1}{(0+h)}=\lim _{h \rightarrow 0} h \cdot \cos \frac{1}{h}=0$

$\lim _{x \rightarrow 0} f(x)=0$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} f(x)=0$

Thus, the given function f(x) is continuous at x = 0.