The value of $k$ for which the function $f(x) = \left\{ \begin{array}{r} \frac{1 - \cos{4x}}{8x^{2}},\ x
eq 0 \ k\ \ \ \ x = 0 \end{array} ight.\$ is continuous at $a = 0$, is

Question

The value of $k$ for which the function $f(x) = \left\{ \begin{array}{r} \frac{1 - \cos{4x}}{8x^{2}},\ x 
eq 0 \ k\ \ \ \ x = 0 \end{array} ight.\$ is continuous at $a = 0$, is

Dr. Abubacker · Accepted Answer

(b)
Given, $f(x) = \left\{ \begin{array}{r}
\frac{1 - \cos{4x}}{8x^{2}},\ x 
eq 0 \
k\ \ \ \ x = 0
\end{array} ight.\$
LHL$= \lim_{x ightarrow 0^{-}}\ \ f(x)$
$= \lim_{h ightarrow 0}\ \frac{1 - \cos{4(0 - h)}}{8{(0 - h)}^{2}}\$
$= \lim_{h ightarrow 0}\frac{1 - \sin{4h}}{8h^{2}}$
$= \lim_{h ightarrow 0}\frac{4\sin{4h}}{16h} = 1$ [by L ‘Hospital’s rule]
Since, $f(x)$ is continuous at $x = 0$
$	herefore\ \ \ \ f(0) =$LHL $\Rightarrow \ \ \ \ k = 1$

Q. The value of $k$ for which the function $f(x) = \left\{ \begin{array}{r} \frac{1 - \cos{4x}}{8x^{2}},\ x \neq 0 \\ k\ \ \ \ x = 0 \end{array} \right.\$ is continuous at $a = 0$, is

Explanation

Reviewed by Dr. Abubacker

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