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

(A) 0

(B) $\frac{1}{2}$

(C) $\frac{1}{4}$

(D) $-\frac{1}{2}$

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Best answer

SOLUTION —

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}=\lim _{x \rightarrow 0} \frac{2 \sin ^2 x / 2}{4(x / 2)^2} x=0$ and $f(0)=k$

$\because$ Function is continuous at $x=0$

$\therefore \quad \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow k=0$

So, The correct option will be (A).

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