Limit Problem

**Sturm88** · September 13th 2008, 02:27 PM

(sqrt7-x)/(49-x^4) as x approaches sqrt7

I multiplied by the conjugate to rationalize the numerator. I also factored the bottom... and I came up with this.

(7-x)/(x^2+7)(x^2-7)

What to do after this?

**o_O** · September 13th 2008, 02:32 PM

$\lim_{x \to \sqrt{7}} \frac{\sqrt{7} - x}{49-x^4}$

$= \lim_{x \to \sqrt{7}} \frac{\sqrt{7} - x}{{\color{red}(7-x^2)}(7+x^2)}$

Try applying the difference of squares formula to the red

**Sturm88** · September 13th 2008, 02:34 PM

haha so obvious... sometimes I don't see these things. Thank you sir!

**o_O** · September 13th 2008, 02:49 PM

Originally Posted by Sturm88

Hey I have a really silly question, but what is a root + a root. like sqrt7 + sqrt 7 in the problem I just posted...

$a + a = 2a$ . This is obvious right?

Imagine $a = \sqrt{7}$

**Sturm88** · September 13th 2008, 02:51 PM

So the answer is 28sqrt7?

**o_O** · September 13th 2008, 02:55 PM

**Sturm88** · September 13th 2008, 03:04 PM

(1/64)-(1/x^2)/(x-8) as x approaches 8

I used difference of squares on the top and I got this

(1/8x+x)(1/8-x)/(x-8) stuck here.

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