evaluating limits

**simsima_1** · February 17th 2008, 11:21 AM

If f(x) = sqrt(x+1) evaluate the following expression:

lim (f(x+h)-f(x))/ h
h->0

so my equation looks something like:

(sqrt(x+h+1)-sqrt(x+1))/h

how do i simplify the square roots? I'm thinking the h's would cancel out somehow in order for this limit not to be undefined but i don't know how that would happen.

**Krizalid** · February 17th 2008, 11:24 AM

Well you're actually contemplating a derivative via definition of derivative.

Next step: multiply top & bottom by $\sqrt {x + h + 1} + \sqrt {x + 1} .$

**simsima_1** · February 17th 2008, 11:34 AM

$<br /> (\sqrt {x + h + 1} - \sqrt {x + 1} )/ h.<br />$

the bottom is just h so how would i multiply the top by itself and the bottom by the numerator?

**wingless** · February 17th 2008, 11:50 AM

$\lim_{h\to 0} \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}$

$\lim_{h\to 0} \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\cdot \frac{\sqrt{x+h+1}+\sqrt{x+1}}{\sqrt{x+h+1}+\sqrt{ x+1}}$

$\lim_{h\to 0} \frac{(x+h+1)-(x+1)}{h(\sqrt{x+h+1}+\sqrt{x+1})}$

$\lim_{h\to 0} \frac{\not h}{\not h(\sqrt{x+h+1}+\sqrt{x+1})}$

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

Put $h=0$ ,

$\frac{1}{2\sqrt{x+1}}$

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