Question

Let $f\left(x\right)=\{\begin{array}{cc}\alpha \left(x\right)\sin \frac{\pi x}{2}& \text{for}x\ne 0\\ 1& \text{for}x=0\end{array}$

where $\alpha \left(x\right)$ is such that $\underset{x\to 0}{lim}|\alpha \left(x\right)|=\infty$.
Then the function $f\left(x\right)$ is continuous at $x=0$ if $\alpha \left(x\right)$ is chosen as :

• A. $\frac{1}{x}$
• B. $\frac{2}{\pi x}$
• C. $\frac{1}{x^2}$
• D. $\frac{2}{\pi x^2}$

Answer 1

The correct option is B $\frac{2}{\pi x}$

For the function to be continuous, limit should exist at $x=0$ and $\underset{x\to 0}{lim}\alpha \left(x\right)\sin \frac{\pi x}{2}=1\Rightarrow \underset{\frac{\pi x}{2}\to 0}{lim}\frac{\sin \frac{\pi x}{2}}{\frac{1}{\alpha \left(x\right)}}=1[\because x\to 0\Rightarrow \frac{\pi x}{2}\to 0]$

Since, $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$
$\therefore \frac{1}{\alpha \left(x\right)}=\frac{\pi x}{2}\Rightarrow \alpha \left(x\right)=\frac{2}{\pi x}$