Question

If the functions f(x), defined below is continuous at x = 0, find the value of k.
$f\left(x\right)=\left\{\begin{array}{cc}\frac{1-\cos 2x}{2x^2},& x<0\\ k,& x=0\\ \frac{x}{\left|x\right|},& x>0\end{array}\right.$

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{c}\frac{1-\cos 2x}{2x^2},\mathrm{x}<0\\ k,x=0\\ \frac{x}{\left|x\right|},x>0\end{array}\right.$


$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}\frac{1-\cos 2x}{2x^2},\mathrm{x}<0\\ k,x=0\\ 1,x>0\end{array}\right.$


We have
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)$

$$\begin{gathered} =\underset{h\to 0}{\lim }\left(\frac{1-\cos 2\left(-h\right)}{2{\left(-h\right)}^2}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{1-\cos 2h}{2h^2}\right) \\ =\frac{1}{2}\underset{h\to 0}{\lim }\left(\frac{2{\sin }^{2}h}{h^2}\right) \\ =\frac{2}{2}\underset{h\to 0}{\lim }\left(\frac{{\sin }^{2}h}{h^2}\right) \\ =\frac{2}{2}\underset{h\to 0}{\lim }{\left(\frac{\sin h}{h}\right)}^2 \\ =1\times 1 \\ =1 \end{gathered}$$

(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }f\left(h\right)$$=\underset{h\to 0}{\lim }\left(1\right)=1$
Also, $f\left(0\right)=k$

If f(x) is continuous at x = 0, then
​$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

$\Rightarrow 1=1=k$

Hence, the required value of k is 1.