Question

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

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

Answer 1

Answer: (B)

$\sqrt{\frac{2}{3}}$

When $x=-1$, the base term reduces to $\frac{2}{3}$. But, when $x=-1$ is substituted in the power, it has $\frac{0}{0}$ form.

Hence, using L'Hospitals rule,

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

$=\underset{x\to -1}{lim}\frac{sin(x+1)}{2(x+1)}$

Now, we still have $\frac{0}{0}$ form. Hence, using L'Hospitals rule again, we get

$\underset{x\to -1}{lim}\frac{sin(x+1)}{2(x+1)}=\underset{x\to -1}{lim}\frac{\frac{d\left(sin\right(x+1\left)\right)}{dx}}{\frac{d\left(2\right(x+1\left)\right)}{dx}}$

$=\underset{x\to -1}{lim}\frac{cos(x+1)}{2}$

$=\frac{cos\left(0\right)}{2}$

$=\frac{1}{2}$

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