Question

Evaluate $\underset{x\to 0}{lim}\frac{2^x-1}{(1+x{)}^{\frac{1}{2}}-1}$

• A. $\log 2$
• B. $\log 4$
• C. $\log 6$
• D. $0$

Answer 1

The correct option is B $\log 4$

If we substitute $x=0$ in the limit, it is $\frac{0}{0}$ type of indeterminate form.
So we can Apply L'Hospitals Rule on $\underset{x\to 0}{lim}\frac{2^x-1}{(1+x{)}^{\frac{1}{2}}-1}$,
$=\underset{x\to 0}{lim}\frac{2^{x}log2}{\frac{1}{2}(1+x{)}^{\frac{-1}{2}}}$

Substitute $x=0$,
$=2log2$
$=log4$