# Trouble rationalising

• Jun 8th 2009, 03:36 PM
Rapid_W
Trouble rationalising
A snippet from a calculus question however i am finding the simplifying of a rationalisation hard.

This $\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$ needs to be rationalised

now i know i multiply both top and bottom by $\frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}$

but I'm having mega difficulty actually doing it and getting an equation that actually looks half decent at the end, if someone could just show me step by step, how to really simplify the equation I would be extremely grateful. Thanks.
• Jun 8th 2009, 04:26 PM
You could simplify the top line to be...

$\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}$

$= \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}$.
• Jun 8th 2009, 04:58 PM
Rapid_W
But if i do that i can't rationalise it.
• Jun 8th 2009, 06:23 PM
VonNemo19
Quote:

Originally Posted by Rapid_W
A snippet from a calculus question however i am finding the simplifying of a rationalisation hard.

This $\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$ needs to be rationalised

now i know i multiply both top and bottom by $\frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}$

but I'm having mega difficulty actually doing it and getting an equation that actually looks half decent at the end, if someone could just show me step by step, how to really simplify the equation I would be extremely grateful. Thanks.

first understand that you can move the h right into the deminator to simplify things a bit. So first do as deadstar showed you. Then whenever you want to ratyionalize either a denominator or numerator, simply multiply both top and bottom by the conjugate of the one you want to rationalize. I assume here you want to rationalize the numerator so that you can remaove that nasty h from the problem.

http://www.mathhelpforum.com/math-he...5aec5a4d-1.gif* $\frac{1}{h}*\frac{(\sqrt{x}+\sqrt{x+h})}{(\sqrt{x} +\sqrt{x+h})}$

= $\frac{x-x+h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}$

= $\frac{1}{x\sqrt{x+h}+\sqrt{x}(x+h)}$

I'll take it one more step for you

as $h\to{0}$ we get

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