Find whether f(x) is continuous or discontinuous at the indicated point $f(x)=\left \{\begin{array}{1 1}\mid x\mid\cos\large\frac{1}{x}, \normalsize& if\;x\neq 0\\0, & if\;x=0\end{array}\right.$at x=0.

Answer 1

Toolbox:

• A function is said to be continuous at $x=a$,if LHL=RHL.
• (i.e) $\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^-}f(x)$

Step 1:

Given : $f(x)=\left \{\begin{array}{1 1}\mid x\mid\cos\large\frac{1}{x}, \normalsize& if\;x\neq 0\\0, & if\;x=0\end{array}\right.$at x=0.

We have $f(x)=\mid x\mid \cos\large\frac{1}{x}$ when $x\neq 0$ and $f(0)=0$

LHL =$\lim\limits_{\large x\to 0^-}f(x)=\lim\limits_{\large x\to 0^-}\mid x\mid \cos\large\frac{1}{x}$

Put $x=0-h$

LHL=$\lim\limits_{\large h\to 0}-(-h).\cos\big(\large\frac{1}{-h}\big)$

But $\cos(-\theta)=\cos\theta$

$\quad\;\;=\lim\limits_{\large h\to 0}(h).\cos\big(\large\frac{1}{h}\big)$

Applying the limits,

$\quad\;\;=0$

Step 2:

Put $x=0+h$

RHL=$\lim\limits_{\large h\to 0} h.\cos\big(\large\frac{1}{h}\big)$

Applying the limits,

$\quad\;\;\;\;=0$

Hence LHL = RHL = 0

Step 3:

Also $f(0)=0$

Therefore $\lim\limits_{\large x\to 0}f(x)=f(0)$

Hence the given function is continuous at $x=0$