Browse Questions

# If $f(x)=\left\{\begin{array}{1 1}\large\frac{1-\cos\lambda x}{x\sin x}&x\neq 0\\\large\frac{1}{2}&x=0\end{array}\right.$ is continuous at $x=0$ then $\lambda$ is

$(a)\;0\qquad(b)\;\pm 1\qquad(c)\;1\qquad(d)\;None\;of\;these$

$\lim\limits_{x\to 0}\large\frac{1-\cos \lambda x}{x\sin x}=$$\lim\limits_{x\to 0}\large\frac{1-\cos \lambda x}{x^2}.\frac{x}{\sin x} \Rightarrow \lim\limits_{x\to 0}\large\frac{1-\cos\lambda x}{x^2}$$=\lim\limits_{x\to 0}\large\frac{2\sin ^2\lambda x/2}{x^2}$
$\Rightarrow \lim\limits_{x\to 0}2\bigg[\large\frac{\sin \lambda x/2}{\lambda x/2}\bigg]^2.\large\frac{\lambda^2}{4}$
$\Rightarrow \large\frac{\lambda^2}{2}$
Also $f(0)=\large\frac{1}{2}$
Since $f$ is continuous at $x=0$
$\large\frac{\lambda^2}{2}=\frac{1}{2}$
$\lambda=\pm 1$
Hence (b) is the correct answer.