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# 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.

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A)
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 $x=0$