Question

Evalaute $\underset{x\to 0}{lim}\frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$

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

Answer 1

The correct option is B $\frac{1}{\log 9}$

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

Using L Hospital's rule

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

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

$=\frac{0-\left(\frac{-1}{2\sqrt{1+0-0}}\right)}{\log 3}=\frac{1}{2\log 3}$

$=\frac{1}{\log 3^2}$

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