Question

sin(πcos^ x)|x^2 . lim x→0 ​

Answer 1

Answer:

The answer is $\pi$

Step-by-step explanation:

$\displaystyle \lim_{x\to 0}\frac{\sin(\pi\cos^2x)}{x^2}=\displaystyle \lim_{x\to 0}\frac{\sin(\pi-\pi\cos^2x)}{x^2}, \because \sin(\pi-\theta)=\sin\theta$

                           $=\displaystyle \lim_{x\to 0}\frac{\sin(\pi(1-\cos^2x))}{x^2}\\=\displaystyle \lim_{x\to 0}\frac{\sin(\pi\sin^2x)}{x^2}\\=\displaystyle \lim_{x\to 0}\frac{\sin(\pi\sin^2x)}{\pi\sin^2 x}\times (\frac{\sin x}{x})^2\times \pi\\=1\times 1^2\times \pi =\pi$

( Using $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$  )