1. ## [SOLVED] Limit question

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

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

2. Hello,

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

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

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$\displaystyle \frac{f(2+h)-f(2)}{h}$ where f(x) = $\displaystyle \sqrt{x}$

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

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

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

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

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

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$\displaystyle \frac{f(2+h)-f(2)}{h}$ where f(x) = $\displaystyle \sqrt{x}$

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