# How do you find the limit of (sqrt(49+h)-7)/h as h approaches 0?

Mar 6, 2015

How: Rationalize the numerator,

(I assume that you have recently begun your study of calculus and do not yet have access to l'Hopital's rule, which is not needed in this problem anyway.)

$\frac{\sqrt{49 + h} - 7}{h} = \frac{\sqrt{49 + h} - 7}{h} \frac{\sqrt{49 + h} + 7}{\sqrt{49 + h} + 7}$

$= \frac{49 + h - 49}{h \left(\sqrt{49 + h} + 7\right)} = \frac{h}{h \left(\sqrt{49 + h} + 7\right)}$

And this last ratio is equal to $\frac{1}{\sqrt{49 + h} + 7}$ for all $h$ except $0$.

${\lim}_{h \rightarrow 0} \frac{\sqrt{49 + h} - 7}{h} = {\lim}_{h \rightarrow 0} \frac{1}{\sqrt{49 + h} + 7} = \frac{1}{\sqrt{49} + 7} = \frac{1}{14}$

The limit as $h \rightarrow 0$ doesn't care what happens when $h$ gets to $0$, only what happen when $h$ is close to $0$.

Another way of expressing this important fact about limits is that if $f \left(x\right) = g \left(x\right)$ for all $x$ 'near' $a$ (except possibly right at $x = a$), then ${\lim}_{x \rightarrow a} f \left(x\right) = {\lim}_{x \rightarrow a} g \left(x\right)$ if either limit exists then they both do and they are equal.

In this problem we used:$\frac{\sqrt{49 + h} - 7}{h} = \frac{1}{\sqrt{49 + h} + 7}$ for all $h$ 'near' $0$ (except right at $h = 0$, where there is no number on the left). We are, therefore justified in asserting that since the limit on the right side of the equation exists, the limit on the other side of the equation must also exist and they are equal.