Question

Find:

${lim}_{n\to 0}\frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}=$

• A. $\frac{1}{\log 3}$
• B. $\log 9$
• C. $\frac{1}{\log 9}$
• D. $\log 3$

Answer 1

Answer: (C)

$\frac{1}{\log 9}$

$\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}\times \frac{\sqrt{1+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x^2}+\sqrt{1-x+x^2}}=\underset{x\to 0}{lim}\frac{(1+x^2)-(1-x+x^2)}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}=\underset{x\to 0}{lim}\frac{x}{(3^x-1)\left(2\right)}$

As this is of $\frac{0}{0}$ form, applying l'optial.

$={lim}_{x\to 0}\frac{1}{2\left(3^x\log 3\right)}$

$=\frac{1}{2\log 3}=\frac{1}{\log 3^2}=\frac{1}{\log 9}\Rightarrow \left(C\right)$