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!

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

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

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

Cancel the obvious common factor and try taking the limit again.

3. 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.

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

FYI New threads should be created for new questions.

5. Mmm, yes I do.

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