If the functions $f (x)$ , defined below is continuous at $x = 0$ , find the value of $k$ :
$f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 2 x}{2 x^{2}}, & x < 0 k, & x = 0 \frac{x}{| x |}, & x > 0 \end{matrix}$

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Solution

$f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 2 x}{2 x^{2}}, & x < 0 K & x = 0 \frac{x}{| x |} & x > 0 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭$
Given, $f (x)$ is continuous at $x = 0$
$\Rightarrow lim x \to 0 f (x) = f (0) = k$
$lim x \to 0^{+} f (x) = lim x \to 0^{+} = \frac{x}{| x |}$
$= lim x \to 0^{+} \frac{x}{x} = 1$
$lim x \to 0^{-} f (x) = lim x \to 0^{-} \frac{1 - cos 2 x}{2 x^{2}}$
(L hospital Rule) $= lim x \to 0^{-} \frac{(+ sin 2 x) ({/ 2}^{2})}{{/ 4}_{2} x}$
$= lim x \to 0^{-} \frac{sin 2 x}{2 x}$
$= 1$
So, $lim x \to 0 f (x) = 1 \Rightarrow K = 1$ .

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