Question

If $f(x)=\displaystyle\lim _{n \rightarrow \infty} \frac{x^{3 n} \sin x+\cos x}{x^{3 n}+2}$, 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

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