Question

Evaluate $\underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}$

• A. $\frac{1}{2\sqrt{x}}$
• B. $\frac{1}{\sqrt{x}}$
• C. $\frac{2}{\sqrt{x}}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2\sqrt{x}}$

Method I
$\underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}$ $\left(\frac{0}{0}form\right)$
$=\underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$
$=\underset{h\to 0}{lim}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}$
$=\underset{h\to 0}{lim}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$
$=\underset{h\to 0}{lim}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}$
Method II
$\underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}$ $\left(\frac{0}{0}form\right)$
$\therefore$ Applying L'Hospital's rule [differentiating numerator and denominator w.r.t. $h$]
$=\underset{h\to 0}{lim}\frac{\frac{1}{2\sqrt{x+h}}-0}{1}=\frac{1}{2\sqrt{x}}$