Question

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

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

Answer 1

Answer: (C)

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


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

= $\underset{x\to 0}{lim}\frac{(1+x^2{)}^{1/2}-[1+(x^2-x){]}^{1/2}}{3^x-1}$

applying L'Hospistals rule

$=\underset{x\to 0}{lim}\frac{\frac{1}{2}(1+x^2{)}^{-1/2}.2x-\frac{1}{2}[1+(x^2-x\left){]}^{-1/2}\right(2x-1)}{3^x\log 3}$

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

$=\frac{1}{log9}$