1. ## limit question

Hey having trouble with this one guys.

$\dfrac{lim}{x \rightarrow -2} : \dfrac{x}{x+2}$

Cheers

2. Definitely the wrong forum, but I ain't the po-lice.

$\frac{x}{x+2} = \frac{1}{1+2/x}$

Wait that did nothing..? One sec.

Oh, approach it from the right and left and see what the deal is.

BTW \lim_{x\rightarrow -2} $= \lim_{x\rightarrow -2}$

Hey having trouble with this one guys.

$\dfrac{lim}{x \rightarrow -2} : \dfrac{x}{x+2}$

Cheers

what about dividing by x to get

$\dfrac{lim}{x \rightarrow -2} : \frac{1}{1+(\frac{2}{x})}$

4. Originally Posted by harish21
what about dividing by x to get

$\dfrac{lim}{x \rightarrow -2} : \frac{1}{1+(\frac{2}{x})}$
What does this accomplish?

5. Originally Posted by Anonymous1
What does this accomplish?

$\dfrac{lim}{x \rightarrow (-2^-)} : \dfrac{x}{x+2} = \infty$

$\dfrac{lim}{x \rightarrow (-2^+)} : \dfrac{x}{x+2} = -\infty$

the limit does not exist!

6. Originally Posted by harish21
$\dfrac{lim}{x \rightarrow (-2^-)} : \dfrac{x}{x+2} = \infty$

$\dfrac{lim}{x \rightarrow (-2^+)} : \dfrac{x}{x+2} = -\infty$

the limit does not exist!
When this happens you average them.

$E[Limit] = \frac{\infty - \infty}{2} = 0.$ So the limit is $0!$

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Don't worry I'm not serious.

7. Thanks guys.

I thought I must have been doing it wrong because I kept getting undefined answers but this explains why