limx→−1 x4+x2+x+1x2−x+11−cos⁡(x+1)(x+1)2 is equal to

A
1
B
$\sqrt{2 / 3}$
C
$\sqrt{3 / 2}$
D
$e^{1 / 2}$

Solution:

We observe that

$\begin{array}{l} \frac{x^{4} + x^{2} + x + 1}{x^{2} - x + 1} \to \frac{2}{3} as x \to - 1 \\ \frac{1 - \cos (x + 1)}{(x + 1)^{2}} \to \frac{0}{0} as x \to - 1 . \\ lim_{x \to - 1} {(\frac{x^{4} + x^{2} + x + 1}{x^{2} - x + 1})}^{1 - \cos (x + 1)} \frac{(x + 1)^{2}}{2 \sin^{2} x + 1} \\ = {(\frac{2}{3})}^{x \to - 1} 1 - \cos (x + 1) \\ lim_{(x + 1)^{2}} = {(\frac{2}{3})}^{lim_{x \to - 1} - \frac{2}{4} {(\frac{x + 1}{2})}^{2}} = {(\frac{2}{3})}^{1 / 2} \end{array}$

$lim_{x \to - 1} {(\frac{x^{4} + x^{2} + x + 1}{x^{2} - x + 1})}^{\frac{1 - \cos (x + 1)}{(x + 1)^{2}}}$ is equal to

Solution:

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