Question

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

• A. $e^{\frac{-1}{12}}$
• B. $\left(2\right)$
• C. $0$
• D. $Doesnotexist$

Answer 1

Answer: (A)

$e^{\frac{-1}{12}}$

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

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

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

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

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

$=e^{\frac{-1}{12}}$ (Ans)