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Q. If $ f(x)=\left\{ \begin{matrix} \frac{1-\cos x}{x}, & x\ne 0 \\ k, & x=0 \\ \end{matrix} \right. $ is continuous at $ x=0, $ then the value of $ k $ is

KEAMKEAM 2008

Solution:

Given, $ f(x)=\left\{ \begin{matrix} \frac{1-\cos x}{x}, & x\ne 0 \\ k, & x=0 \\ \end{matrix} \right. $ $ \underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{x} $
$=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\frac{x}{2}}{4\left( \frac{x}{2} \right)}.x=0 $ and $ f(0)=k $
$ \because $ Function is continuous at $ x=0 $ .
$ \therefore $ $ \underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\Rightarrow k=0 $