Question

lim x tends to zero sin(πcos^2x X )/x^2

Answer 1

here is the answer for your question

Answer 2

$\bold{\pi}$ is the value of $\bold{\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}}$

Given:

$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}$

To find:

Value of  $\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}} = ?$

Solution:

First let us convert $\cos ^{2} x \text { into }\left(1-\sin ^{2} x\right)$after converting we get

$\begin{array}{l}{\lim _{x \rightarrow 0} \frac{\sin \left(\pi\left(1-\sin ^{2} x\right)\right)}{x^{2}}} \\ {\lim _{x \rightarrow 0} \frac{\sin \left(\left(\pi-\pi \sin ^{2} x\right)\right)}{x^{2}}}\end{array}$

Converting $\pi-\pi \sin ^{2} x \text { into } \pi \sin ^{2} x\lim _{x \rightarrow 0} \frac{\sin \left(\left(\pi \sin ^{2} x\right)\right)}{x^{2}}$

As we know and for those who don’t know the value of πsin^2 x always tends to zero, therefore we get the value of  

$\lim _{x \rightarrow 0} \frac{\left(\left(\pi \sin ^{2} x\right)\right)}{x^{2}}$

Which can be also be written as  

$\lim _{x \rightarrow 0} \frac{\left(\pi . \sin ^{2} x\right)}{x^{2}}=\lim _{x \rightarrow 0} \pi \cdot \frac{\sin ^{2} x}{x^{2}}$

Therefore after putting x = 0 we get   $\pi . \frac{\sin ^{2} x}{x^{2}}=\pi \cdot \frac{0}{0}=\pi$

Hence, the answer is $\bold{\pi.}$