Limit

• Sep 7th 2009, 06:57 PM
Nightasylum
Limit
OK I am getting a different answer then the book.
Find the limit
x------>4
square root(x-3) - 1 / x-4
X-3 is the only part in the fraction that is square rooted.
I know you can bring it down and be lefted with
1/square (x-3) which gives you lim X=1 back the books says its one half so im doing something wrong .
• Sep 7th 2009, 06:59 PM
Chris L T521
Quote:

Originally Posted by Nightasylum
OK I am getting a different answer then the book.
Find the limit
x------>4
square root(x-3) - 1 / x-4
X-3 is the only part in the fraction that is square rooted.
I know you can bring it down and be lefted with
1/square (x-3) which gives you lim X=1 back the books says its one half so im doing something wrong .

$\displaystyle \lim_{x\to4}\frac{\sqrt{x-3}-1}{x-4}=\lim_{x\to4}\frac{\sqrt{x-3}-1}{x-4}\cdot\frac{\sqrt{x-3}+1}{\sqrt{x-3}+1}$ $\displaystyle =\lim_{x\to4}\frac{x-3-1}{\left(x-4\right)\left(\sqrt{x-3}+1\right)}=\lim_{x\to4}\frac{1}{\sqrt{x-3}+1}$

I'm sure you can take it from here...
• Sep 7th 2009, 07:10 PM
Nightasylum
Ok so you acually rationalize the whole numerator? I was just doing the square root part.
Also can you help with
X---->-5
x^3+125/x+5
• Sep 7th 2009, 07:15 PM
Chris L T521
Quote:

Originally Posted by Nightasylum
Ok so you acually rationalize the whole numerator? I was just doing the square root part.
Also can you help with
X---->-5
x^3+125/x+5

You will need to use the formula for factoring the sum of two cubes:

$\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)$
• Sep 7th 2009, 07:28 PM
Nightasylum
(x+5)(x^2-5x+5^2)/(x+5) = x^2-5x+5^2 = 75
I think I did that right I need to remember that formula thanks alot.
• Sep 7th 2009, 07:35 PM
Chris L T521
Quote:

Originally Posted by Nightasylum
(x+5)(x^2-5x+5^2)/(x+5) = x^2-5x+5^2 = 75
I think I did that right I need to remember that formula thanks alot.

(Yes)

$\displaystyle \lim_{x\to-5}\frac{x^3+125}{x+5}=75$