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. ${\left(\frac{2}{3}\right)}^{1/2}$
• C. ${\left(\frac{3}{2}\right)}^{1/2}$
• D. $e^{1/2}$

Answer 1

Answer: (B)

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