Question

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=$

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

Answer 1

Answer: (A)

$-\pi$

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=\frac{\sin \left(\pi {\cos }^2\right(0\left)\right)}{0}=\frac{sin\left(\pi \right)}{0}=\frac{0}{0}\left(\text{indeterminate form}\right)$

According to L'Hospital's rule:

Suppose that we have the following cases

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}$

where $a$ can be any real number, infinity or negative infinity. In these cases we have,

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$ $where,f^{\prime }\left(x\right)\text{is derivative of f(x) with respect to x}$

Hence, in given question we can apply L'Hospital's rule.

Here,

$f\left(x\right)=\sin \left(\pi {\cos }^{2}x\right)andg\left(x\right)=x^2$

and

$f^{\prime }\left(x\right)=\cos \left(\pi {\cos }^{2}x\right)(-2\pi \cos x\sin x)$

or

$f^{\prime }\left(x\right)=-\pi \sin \left(2x\right)\cos \left(\pi {\cos }^{2}x\right)\because 2\sin x\cos x=\sin \left(2x\right)$

and

$g^{\prime }\left(x\right)=2x$

Now,

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=\underset{x\to 0}{lim}\frac{-\pi \sin \left(2x\right)\cos \left(\pi {\cos }^{2}x\right)}{2x}=-\pi \underset{x\to 0}{lim}(\frac{\sin 2x}{2x})\underset{x\to 0}{lim}(\cos \left(\pi {\cos }^{2}x\right))$

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=-\pi \underset{x\to 0}{lim}\cos \left(\pi {\cos }^{2}x\right)\because \underset{t\to 0}{lim}\frac{\sin t}{t}=1$

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=-\pi \cos \left(\pi {\cos }^{2}0\right)$

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=-\pi \cos \left(\pi \right)\because \cos 0=1$

$\underset{x\to 0}{lim}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=\pi \because \cos \pi =-1$