Answer
$\lim_{h\to0}\dfrac{\sqrt{1+h}-1}{h}=\dfrac{1}{2}$
Work Step by Step
$\lim_{h\to0}\dfrac{\sqrt{1+h}-1}{h}$
Try to evaluate the limit applying direct substitution:
$\lim_{h\to0}\dfrac{\sqrt{1+h}-1}{h}=\dfrac{\sqrt{1+0}-1}{0}=\dfrac{0}{0}$ Indeterminate form
Rationalize the numerator of the function and simplify:
$\lim_{h\to0}\dfrac{\sqrt{1+h}-1}{h}=\lim_{h\to0}\dfrac{\sqrt{1+h}-1}{h}\cdot\dfrac{\sqrt{1+h}+1}{\sqrt{1+h}+1}=...$
$...=\lim_{h\to0}\dfrac{(\sqrt{1+h})^{2}-1^{2}}{h(\sqrt{1+h}+1)}=\lim_{h\to0}\dfrac{1+h-1}{h(\sqrt{1+h}+1)}=...$
$...=\lim_{h\to0}\dfrac{h}{h(\sqrt{1+h}+1)}=\lim_{h\to0}\dfrac{1}{\sqrt{1+h}+1}=...$
Try to evaluate the limit using direct substitution again:
$\lim_{h\to0}\dfrac{1}{\sqrt{1+h}+1}=\dfrac{1}{\sqrt{1+0}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}$
