Question

$\underset{x\to -1}{lim}\frac{x+1}{2-\sqrt{4+x++x^2}}$=

• A. $\frac{1}{4}$
• B. $4$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$4$

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

this is of the form $\%$, so we use $L^{\prime }$ hospital rule

if $\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}f\left(a\right)=g\left(a\right)=0$

Then ${lim}_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}={lim}_{x\to a}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

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

$\Rightarrow \underset{x\to -1}{lim}-\frac{2\sqrt{4+x+x^2}}{(1+2x)}$

$\Rightarrow \frac{-2\sqrt{4-1+1}}{1+2(-1)}$
$\Rightarrow \frac{-2\times 2}{-1}$

$\Rightarrow 4$

$\therefore \underset{x\to -1}{lim}\frac{x+1}{2-\sqrt{4+x+x^2}}=4$