• September 11th 2008, 06:21 PM
Toaster
Ok, so here is the limit question that has been giving me some difficulty:

The lim of x+2 divided by the square root of 6 + x minus the square root of 2 - x as x approaches -2

(Sorry, if I don't know how to put it in symbols otherwise I would >//<; )

Also, any general tips on finding the proper way to "play with"/solve limits would be much appreciated. Thanks for any help in advance!
• September 11th 2008, 06:55 PM
mr fantastic
Quote:

Originally Posted by Toaster
Ok, so here is the limit question that has been giving me some difficulty:

The lim of x+2 divided by the square root of 6 + x minus the square root of 2 - x as x approaches -2

(Sorry, if I don't know how to put it in symbols otherwise I would >//<; )

Also, any general tips on finding the proper way to "play with"/solve limits would be much appreciated. Thanks for any help in advance!

$\lim_{x \rightarrow -2} \frac{x+2}{\sqrt{6+x} - \sqrt{2-x}}$.

Multiply the expression by $1 = \frac{ \sqrt{6+x} + \sqrt{2-x} }{\sqrt{6+x} + \sqrt{2-x}}$ and simplify the denominator. You should get:

$\lim_{x \rightarrow -2} \frac{(x+2)(\sqrt{6+x} + \sqrt{2-x})}{4 + 2x}$.

Cancel the obvious common factor and try taking the limit again.
• September 11th 2008, 07:04 PM
Toaster
Ah! Thank you! I see I shouldn't have multiplied the top out, do you think you could help me with another question?

The limit of x-8 divided by (the cube root of x) - 2 as x approaches 8.
• September 11th 2008, 07:09 PM
mr fantastic
Quote:

Originally Posted by Toaster
Ah! Thank you! I see I shouldn't have multiplied the top out, do you think you could help me with another question?

The limit of x-8 divided by (the cube root of x) - 2 as x approaches 8.

Hint: $x - 8 = (x^{1/3})^3 - 2^3$. You know how to factorise a difference of two cubes, right ....?

FYI New threads should be created for new questions.
• September 11th 2008, 07:12 PM
Toaster
Mmm, yes I do.

Thanks for the hint and the heads-up. ^_^