Question

Let $\varphi (x)=\frac{1-cos\lambda }{xsinx},x\ne 0$ and $\varphi (0\frac{1}{2}$ . If $\varphi (x)$ is continuous at $x$ = 0, then $\lambda$ is

• A. 0
• B. $\pm 1$
• C. 2
• D. none of these.

Answer 1

Answer: (B)

${\lim }_{x\to 0}\varphi \left(x\right)={\lim }_{x\to 0}\frac{1-\cos \lambda x}{x\sin x}={\lim }_{x\to 0}\frac{1-\cos \lambda x}{x^2}$
= ${\lim }_{x\to 0}\frac{2si{n}^2\frac{\lambda x}{2}}{x^2}$
= ${\lim }_{x\to 0}2{\left(\frac{\frac{\sin \frac{\lambda x}{2}}{\lambda x}}{2}\right)}^2\frac{{\lambda }^2}{4}=\frac{{\lambda }^2}{2}$
Also $\varphi (0)=\frac{1}{2}$
Since $\varphi (x)$ is continuous at $x$ = 0
$\therefore$ ${\lim }_{x\to 0}\varphi \left(x\right)=\varphi (0)$
$\Rightarrow \frac{{\lambda }^2}{2}=\frac{1}{2}\Rightarrow {\lambda }^2=1\Rightarrow \lambda =\pm 1$