# Prove the following identities.

Question:

$f(x)=\left\{\begin{array}{l}|x-a| \sin \frac{1}{x-a}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=a\end{array}\right.$ at $\mathrm{x}=\mathrm{a}$

Solution:

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

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

$=\lim _{h \rightarrow 0}|a-h-a| \cdot \sin \frac{1}{a-h-a}=\lim _{h \rightarrow 0} h \cdot \sin \frac{1}{-h}$

$=\lim _{h \rightarrow 0}-h \cdot \sin \frac{1}{h} \quad[\because \sin (-\theta)=-\sin \theta]$

$=0 \times[$ a number oscillating between $-1$ and 1$]$

$=0$

$\lim _{x \rightarrow a^{+}} f(x)=|x-a| \sin \frac{1}{x-a}$

$=\lim _{h \rightarrow 0}|a+h-a| \cdot \sin \frac{1}{a+h-a}=\lim _{h \rightarrow 0} h \cdot \sin \frac{1}{h}$

$=0 \times[$ a number oscillating between $-1$ and 1$]$

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

Now, as

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

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