1. ## [SOLVED] Limit question

limit as h goes to 0 of $\frac{f(2+h)-f(2)}{h}$ where f(x) = $\sqrt{x}$

It looks similar to differentiation from first principles, but I can't see how to use that...

2. Hello,

Remember that $(a-b)(a+b)=a^2-b^2$

This is a common trick : multiply by $1=\frac{f(2+h)+f(2)}{f(2+h)+f(2)}$

The numerator will simplify with the denominator, and what's left is no longer undefined

3. Originally Posted by PTL
limit as h goes to 0 of $\frac{f(2+h)-f(2)}{h}$ where f(x) = $\sqrt{x}$

It looks similar to differentiation from first principles, but I can't see how to use that...
$\lim_{h \to 0} \frac{\sqrt{2 + h} - \sqrt{2}}{h}$

Rationalize the numerator:
$= \lim_{h \to 0} \frac{\sqrt{2 + h} - \sqrt{2}}{h} \cdot \frac{\sqrt{2 + h} + \sqrt{2}}{\sqrt{2 + h} + \sqrt{2}}$

$= \lim_{h \to 0} \frac{2 + h - 2}{h(\sqrt{2 + h} + \sqrt{2})}$

$= \lim_{h \to 0} \frac{h}{h(\sqrt{2 + h} + \sqrt{2})}$

$= \lim_{h \to 0} \frac{1}{\sqrt{2 + h} + \sqrt{2}}$

Now you can evaluate the limit by direct substitution.

01

EDIT: too slow...

4. Hi,
Originally Posted by PTL
limit as h goes to 0 of $\frac{f(2+h)-f(2)}{h}$ where f(x) = $\sqrt{x}$

It looks similar to differentiation from first principles, but I can't see how to use that...
If you are supposed to know that $f$ is differentiable at $x=2$ then you can claim that
$\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=f'(2)$.
and all you have to do is compute $f'(2)$...