Question

$\underset{x\to -1}{lim}{\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)}^{\frac{1-\cos (x+1)}{(x+1{)}^2}}$ is equal to

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

Answer 1

Answer: (B)

$(\frac{2}{3}{)}^{\frac{1}{2}}$

$\underset{x\to -1}{lim}{\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)}^{\frac{1-\cos (x+1)}{(x+1{)}^2}}$
$=\underset{x\to -1}{lim}{\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)}^{\underset{x\to -1}{lim}\frac{1-\cos (x+1)}{(x+1{)}^2}}$
$={\left(\frac{2}{3}\right)}^{\underset{x\to -1}{lim}\frac{2{\sin }^2\frac{(x+1)}{2}}{(x+1{)}^2}}$
$={\left(\frac{2}{3}\right)}^{\underset{x\to -1}{lim}\frac{2{\sin }^2\frac{(x+1)}{2}}{4\frac{(x+1{)}^2}{4}}}$
$={\left(\frac{2}{3}\right)}^{\frac{1}{2}}$