# Thread: Another issue with rationalizing

1. ## Another issue with rationalizing

I ran into another rationalizing problem and it is messing me up. it is f(x) = 1/sqrtx

so when I take my f(a+h) - f(a) I get {[1/sqrt(x+h)] - [1/sqrtx]}/h but inlike the question before that dealt with indeterminate form I am finding rationalizing is getting a bit messy. I end up with [(sqrt(x+h) + sqrtx)/h]/h

I will get this algebra someday. Haha

2. ## Re: Another issue with rationalizing

I'm not sure what your question is or exactly HOW you "rationalized". The point of rationalizing is that (a- b)(a+ b)= a^2- b^2 so if a or b involved a square root, we will have gotten rid of it by squaring.

I could see two ways of approaching $\displaystyle \frac{1}{\sqrt{x+h}}- \frac{1}{\sqrt{x}}$.

a) Multiply by $\displaystyle \frac{\frac{1}{\sqrt{x+h}}+ \frac{1}}{\sqrt{x}}{\frac{1}{\sqrt{x+h}}- \frac{1}{\sqrt{x}}}$ to get
$\displaystyle \frac{\frac{1}{x+ h}- \frac{1}{x}}{\frac{1}{\sqrt{x+h}}+ \frac{1}{\sqrt{x}}}$
so the numerator can be written as $\displaystyle \frac{x- x- h}{x(x+ h)}= \frac{-h}{x(x+h)}$ and then, divided by h,
$\displaystyle \frac{-1}{x(x+ h)}$ so that, with the denominator, we have
$\displaystyle \frac{-1}{(x(x+h)(\frac{1}{\sqrt{x+h}}+ \frac{1}{\sqrt{x}}}$

The limit, as h goes to 0 is $\displaystyle \frac{-1}{x^2(\frac{2}{\sqrt{x}})}= \frac{-\sqrt{x}}{2x^2}= -\frac{1}{2}x^{-3/2}$

Personally, I would combine the fractions first: $\displaystyle \frac{1}{\sqrt{x+h}}- \frac{1}{\sqrt{x}}= \frac{\sqrt{x}- \sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}$
Now, multiply numerator and denominator by $\displaystyle \frac{\sqrt{x}+ \sqrt{x+h}}$ to get
$\displaystyle \frac{x- x- h}{(\sqrt{x}+ \sqrt{x+ h})(\sqrt{x+h}\sqrt{x})}= \frac{-h}{(\sqrt{x}+ \sqrt{x+ h})(\sqrt{x+h}\sqrt{x})}$

Dividing by h and taking the limit as h goes to 0 gives the same as before.

3. ## Re: Another issue with rationalizing

Originally Posted by Jonroberts74
I ran into another rationalizing problem and it is messing me up. it is f(x) = 1/sqrtx
so when I take my f(a+h) - f(a) I get {[1/sqrt(x+h)] - [1/sqrtx]}/h but inlike the question before that dealt with indeterminate form I am finding rationalizing is getting a bit messy. I end up with [(sqrt(x+h) + sqrtx)/h]/h
To make the "algebra" less messy write it as:
$\displaystyle f(x)=\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}$.

4. ## Re: Another issue with rationalizing

Thank you, I see what I was doing wrong.