If f(x)= c 1- x x , x 0 k , x=0 . is continuous at x=0, then k =

MCQ

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

✓
$0$
B
$\frac{1}{2}$
C
$\frac{1}{4}$
D
$-\frac{1}{2}$

✓

Answer

Correct option: A.

$0$

(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}=\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{x}{4}$
$\therefore \quad f(0)=2(1)(0)=0$
Alternate method:
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\rightarrow f (0)=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \sin x=0$

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