Q67 / 170 · Manipal (MET) 2015 · Mathematics
$\lim _{x \rightarrow-1}\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)^{\frac{1-\cos (x+1)}{(x+1)^2}}$ is equal to
- A$1$
- Correct answerB$\sqrt{\frac{2}{3}}$
- C$\sqrt{\frac{3}{2}}$
- D$e^{1 / 2}$
Solution
We observe that
$\frac{x^4+x^2+x+1}{x^2-x+1} \rightarrow \frac{2}{3}$ as $x \rightarrow-1$
But $\quad \frac{1-\cos (x+1)}{(x+1)^2} \rightarrow \frac{0}{0}$ as $x \rightarrow-1$
$\therefore \quad \lim _{x \rightarrow-1}\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)^{\frac{1-\cos (x+1)}{(x+1)^2}}$
$=\left(\frac{2}{3}\right) x^{\lim _{x \rightarrow-1} \frac{1-\cos (x+1)}{(x+1)^2}}=\left(\frac{2}{3}\right)^{\lim _{x \rightarrow-1} \frac{2 \sin ^2\left(\frac{x+1}{2}\right)}{4\left(\frac{x+1}{2}\right)^2}}=\left(\frac{2}{3}\right)^{1 / 2}$
Hence, the correct answer is option: 'B'.