# Thread: Limit question, likely a simple mistake on my part.

1. ## Limit question, likely a simple mistake on my part.

Hey

Question:
Evaluate the following limit: $\displaystyle \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x}$

Solution:
I tried multiplying by the conjugate, but that gives me: $\displaystyle \frac {x - 9}{(9 - x)(\sqrt {x} + 3)}$

and I really don't know what to do from here. The top now gives 0, and I can't factor it out using the bottom. Help?

2. Originally Posted by Kakariki
Hey

Question:
Evaluate the following limit: $\displaystyle \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x}$

Solution:
I tried multiplying by the conjugate, but that gives me: $\displaystyle \frac {x - 9}{(9 - x)(\sqrt {x} + 3)}$

and I really don't know what to do from here. The top now gives 0, and I can't factor it out using the bottom. Help?
Your method is correct! $\displaystyle \frac {x - 9}{(9 - x)(\sqrt {x} + 3)} = \frac {x - 9}{(-(x-9))(\sqrt {x} + 3)} = \frac{1}{-(\sqrt {x} + 3)} = \frac{-1}{6}$

OR,

dont multiply. factorize the denominator:

$\displaystyle \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(x-9)} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(\sqrt{x} - 3)(\sqrt{x} + 3)}= \lim_{x\to9} \frac{1}{\sqrt{x} + 3} = \frac{-1}{6}$

3. Originally Posted by harish21
Your method is correct! $\displaystyle \frac {x - 9}{(9 - x)(\sqrt {x} + 3)} = \frac {x - 9}{(-(x-9))(\sqrt {x} + 3)} = \frac{1}{-(\sqrt {x} + 3)} = \frac{-1}{6}$

OR,

dont multiply. factorize the denominator:

$\displaystyle \lim_{x\to9} \ \frac {\sqrt{x} - 3}{9 - x} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(x-9)} = \lim_{x\to9} \ \frac {\sqrt{x} - 3}{-(\sqrt{x} - 3)(\sqrt{x} + 3)}= \lim_{x\to9} \frac{1}{\sqrt{x} + 3} = \frac{-1}{6}$
Oh dear, that is an easy thing to do. I just factored the negative out of the 9 - x and then got - 1/6. that's really easy
Thanks!!!
BTW, if that is Iron maiden as your avatar, you are awesome. If not, you are still awesome for helping me

4. Originally Posted by Kakariki
Oh dear, that is an easy thing to do. I just factored the negative out of the 9 - x and then got - 1/6. that's really easy
Thanks!!!
BTW, if that is Iron maiden as your avatar, you are awesome. If not, you are still awesome for helping me
Haha! Yes, Thats Iron Maiden.