Question

If $f(x)=\underset{n\to \infty }{\lim }\frac{x^{3n}\sin x+\cos x}{x^{3n}+2}$, then find the value of $\left[f\left(\frac{\pi }{6}\right)+f\left(\frac{\pi }{3}\right)\right]$ taking $\sqrt{3}=1.73$, where $[]$represents the greatest integer function.

Answer 1

Answer: 1

If $|x|<1$, then
$f(x)=\underset{n\to \infty }{\lim }\left(\frac{x^{3n}\sin x+\cos x}{x^{3n}+2}\right)$
$=\frac{\cos x}{2}$... $\left[\underset{n\to \infty }{\lim }{x}^n=0\text{ if }|x|<1\right]$
$\Rightarrow f\left(\frac{\pi }{6}\right)=\frac{1}{2}\cos \frac{\pi }{6}=\frac{\sqrt{3}}{4}$
If $|x|>1$ then $f(x)=\underset{n\to \infty }{\lim }\left(\frac{x^{3n}\sin x+\cos x}{x^{3n}+2}\right)$
$=\underset{n\to \infty }{\lim }\left(\frac{\sin x+x^{-3n}\cos x}{1+2x^{-3n}}\right)$
$=\sin x$
$\Rightarrow f\left(\frac{\pi }{3}\right)=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$
$f\left(\frac{\pi }{6}\right)+f\left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}=1.2975$
$\Rightarrow \left[f\left(\frac{\pi }{6}\right)+f\left(\frac{\pi }{3}\right)\right]=1$