Question

The value of $limx\to \infty x^{2}sin\left(\iota n\sqrt{cos\frac{\pi }{x}}\right)$ is

• A. $-\frac{{\pi }^2}{2}$
• B. $-\frac{{\pi }^2}{4}$
• C. $\frac{{\pi }^2}{2}$
• D. $\frac{{\pi }^2}{4}$

Answer 1

The correct option is B $-\frac{{\pi }^2}{4}$

$\underset{x\to \infty }{lim}{x}^2\left(\ln \sqrt{\cos \frac{\pi }{x}}\right)$

$\underset{y\to 0}{lim}\left\{\frac{\sin \left[\ln \sqrt{\cos \pi y}\right]}{y^2}\right\}....[x+\frac{1}{y}]$

This is $\left(\frac{0}{0}\right)$ form

By $L-$Hospital Rule

$\underset{y\to 0}{lim}-\frac{\cos \left[\ln \sqrt{\cos \pi y}\right]\left(\pi \right)\sin \pi y}{2\left(2y\right)\sqrt{\cos \pi y}\sqrt{\cos \pi y}}$

$\frac{-{\pi }^2}{4}\underset{y\to 0}{lim}\left[\left\{\frac{\cos \left(\ln \sqrt{\cos \pi y}\right)}{\left(\cos \pi y\right)}\right\}\left(\frac{\sin \pi y}{\pi y}\right)\right]$

$=(-\frac{{\pi }^2}{4})$