Question

What is $\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}$ equal to?

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $-\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{2}$

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

$=\underset{x\to 0}{lim}\frac{x+1-1}{x\cdot (\sqrt{x+1}+1)}=\underset{x\to 0}{lim}\frac{1}{(\sqrt{x+1}+1)}$

$=\frac{1}{2}$
Hence, B is correct.