# Limit

• February 27th 2011, 06:26 AM
stripe501
Limit
I that the limit as x->2 of f(x)=(2x^2-8)/(x-2) is 8. But i don't understand how i get there, please help, i'm not really sure what i'm supposed to do

*Has been edited to show the original question
*Solved using factorising
• February 27th 2011, 06:37 AM
FernandoRevilla
Quote:

Originally Posted by stripe501
I that the limit as x-> 0 of f(x)=(2x^2-8)/(x^2-4) is 8. But i don't understand how i get there, please help, i'm not really sure what i'm supposed to do

Your function is continuous at $x=0$ so, the limit is $f(0)=2$ .
• February 27th 2011, 06:39 AM
skeeter
Quote:

Originally Posted by stripe501
I that the limit as x-> 0 of f(x)=(2x^2-8)/(x^2-4) is 8. no, it's not ... did you post the limit correctly?

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• February 27th 2011, 06:40 AM
stripe501
Never mind, I figured it out :D
• February 27th 2011, 06:40 AM
stripe501
Yeah, I posted it wrong :/ thats why i couldn't get it