Limit question, likely a simple mistake on my part.

**Kakariki** · March 29th 2010, 10:29 PM

Hey

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

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

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?

**harish21** · March 29th 2010, 10:37 PM

Originally Posted by Kakariki

Hey

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

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

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

OR,

dont multiply. factorize the denominator:

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

**Kakariki** · March 29th 2010, 10:42 PM

Originally Posted by harish21

Your method is correct! $\frac {x - 9}{(9 - x)(\sqrt {x} + 3)} = \frac {x - 9}{(-(x-9))(\sqrt {x} + 3)} = \frac{1}{-(\sqrt {x} + 3)} = \frac{-1}{6}$

OR,

dont multiply. factorize the denominator:

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

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

**harish21** · March 29th 2010, 10:49 PM

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.

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