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

A
1
B
${(\frac{2}{3})}^{\frac{1}{2}}$
C
${(\frac{3}{2})}^{\frac{1}{2}}$
D
$e^{\frac{1}{2}}$

Solution:

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

$= lim_{x \to - 1} {(\frac{x^{4} + x^{2} + x + 1}{x^{2} - x + 1})}^{lim_{x \to - 1} \frac{1 - \cos (x + 1)}{(x + 1)^{2}}} = {(\frac{2}{3})}^{lim_{x \to - 1}} \frac{2 \sin^{2} \frac{(x + 1)}{2}}{(x + 1)^{2}} = {(\frac{2}{3})}^{lim_{x \to - 1} \frac{2 \sin^{2} \frac{(x + 1)}{2}}{4 \frac{(x + 1)^{2}}{4}}} = {(\frac{2}{3})}^{\frac{1}{2}}$

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

Solution:

Related content