# How do you evaluate the limit (sqrt(x+1)-2)/(x-3) as x approaches 3?

Aug 4, 2016

$= \frac{1}{4}$

#### Explanation:

${\lim}_{x \to 3} \frac{\sqrt{\left(x + 1\right)} - 2}{x - 3}$

with $x = 3 + h , h = x - 3 , 0 < \left\mid h \right\mid \text{<<} 1$

becomes ${\lim}_{h \to 0} \frac{\sqrt{3 + h + 1} - 2}{3 + h - 3}$

${\lim}_{h \to 0} \frac{1}{h} \cdot \left(\sqrt{4 + h} - 2\right)$

${\lim}_{h \to 0} \frac{1}{h} \cdot \left(2 \sqrt{1 + \frac{h}{4}} - 2\right)$

by Taylor expansion

${\lim}_{h \to 0} \frac{1}{h} \cdot \left(2 \left(1 + \frac{1}{2} \cdot \frac{h}{4} + m a t h c a l \left(O\right) \left({h}^{2}\right)\right) - 2\right)$

${\lim}_{h \to 0} \frac{1}{h} \cdot 2 \left(\frac{1}{2} \cdot \frac{h}{4} + m a t h c a l \left(O\right) \left({h}^{2}\right)\right)$

${\lim}_{h \to 0} \frac{1}{h} \cdot \left(\frac{h}{4} + m a t h c a l \left(O\right) \left({h}^{2}\right)\right)$

${\lim}_{h \to 0} \frac{1}{4} + m a t h c a l \left(O\right) \left(h\right) = \frac{1}{4}$

You can see clearly that the limit is the same as x approaches 3 on both sides as the sign of h is in the error term.