# having algebra issues again

• Sep 21st 2013, 11:37 AM
Jonroberts74
having algebra issues again
hello,

I had a recent post in which I explained I am new, I am trying to teach myself mathematics. I am having algebra issues again with a limit problem of the indeterminate form.

Attachment 29224

thank you
• Sep 21st 2013, 11:49 AM
Plato
Re: having algebra issues again
Quote:

Originally Posted by Jonroberts74
I had a recent post in which I explained I am new, I am trying to teach myself mathematics. I am having algebra issues again with a limit problem of the indeterminate form.
Attachment 29224

You need to learn some algebra first.

Rewrite it as $\frac{{\sqrt {x + 1} - \sqrt {2x + 1} }}{{\sqrt {3x + 4} - \sqrt {2x + 4} }} = - \frac{{\sqrt {3x + 4} + \sqrt {2x + 4} }}{{\sqrt {x + 1} + \sqrt {2x + 1} }}$.
• Sep 21st 2013, 12:03 PM
Jonroberts74
Re: having algebra issues again
I've been reviewing my algebra books, I couldn't find anything that showed to make it a negative reciprocal. can you explain a little bit? thank you
• Sep 21st 2013, 12:51 PM
Plato
Re: having algebra issues again
Quote:

Originally Posted by Jonroberts74
I've been reviewing my algebra books, I couldn't find anything that showed to make it a negative reciprocal. can you explain a little bit? thank you

You rationalize by multiplying the original expression by
$\left( {\frac{{\sqrt {x + 1} + \sqrt {2x + 1} }}{{\sqrt {3x + 4} + \sqrt {2x + 4} }}\frac{{\sqrt {3x + 4} + \sqrt {2x + 4} }}{{\sqrt {x + 1} + \sqrt {2x + 1} }}} \right)$

Note that you have the "sum and difference" of radicals in the numerator and denominator.
• Sep 21st 2013, 01:01 PM
votan
Re: having algebra issues again
Quote:

Originally Posted by Jonroberts74
I've been reviewing my algebra books, I couldn't find anything that showed to make it a negative reciprocal. can you explain a little bit? thank you

What Plato showed you is to multiply and divide by the conjugate of the numerator, then multiply and devide by the conjugate of the denominator. Then use the generic formula a^2 - b^2 = (a - b)(a+ b) and simplify. Oh, the conjjugate (a-b) is (a+b).

Replace x by 0 you will get -2.
• Sep 21st 2013, 01:34 PM
Jonroberts74
Re: having algebra issues again
Thank you very much. I understand now, I don't know why I did not see it before. It can be overwhelming teaching yourself math.
• Sep 21st 2013, 07:09 PM
Jonroberts74
Re: having algebra issues again
I have another question relating back to this. I spent the last few hours reading about radicals and rationalizing them. In this problem, you multiple numerator and denominator by their respective conjugate then by each others conjugate in order to solve. In another problem I had to rationalize the numerator but not the denominator because it had a radical up top and a rational number on bottom. This left me with a radical on bottom. Previously, I have only rationalized the denominator.

How do I know which Process of the three I need to use? Or for limits do I rationalize each part that contains a radical.

Thank you very much, maybe one day I will be helping people learn math.
• Sep 21st 2013, 11:35 PM
ibdutt
Re: having algebra issues again
• Sep 22nd 2013, 11:38 AM
Jonroberts74
Re: having algebra issues again
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
• Sep 22nd 2013, 02:42 PM
votan
Re: having algebra issues again
Quote:

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

It is not really messy. Convert the difference to a common denominator then rationalize the numerator. Take the limit h = 0 and that will be it.