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

Question

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

Dr. Abubacker · Accepted Answer

(c)
Since, $\lim_{x \rightarrow 0}\ \ f(x) = f(0)$
$\Rightarrow \ \ \ \lim_{x \rightarrow 0}\ \frac{1 - \cos x}{x^{2}} = k$
$\Rightarrow \ \ \ \lim_{x \rightarrow 0}\ \frac{–( - \sin x)}{2x} = k$ [using L ‘Hospital’s rule]
$\Rightarrow \ \ \ \ \frac{1}{2}\lim_{x \rightarrow 0}\ \frac{\sin x}{x} = k\ \ \ \ \  \Rightarrow \ \ \ \ \ k = \frac{1}{2}$

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

Explanation

Reviewed by Dr. Abubacker

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