Find Limit of a Function Homework | Wolfram Alpha

[azncocoluver]

Homework Statement

http://www4b.wolframalpha.com/Calculate/MSP/MSP10351a01a355263db6f2000030459ebe888feb73?MSPStoreType=image/gif&s=10&w=131&h=41
I know the answer, but I don't know how to get it.

Homework Equations

If picture doesn't work: limit x->1 (x-1)/((square root(x+3) - 2)

The Attempt at a Solution

I know that when you plug in 1, the answer becomes 0/0, which means I need to factor. I don't know how to factor this equation, though.

 

[CJ256]

http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:1,SMB02FSMB03x-1SMB10SMB02RSMB03x+3SMB02rSMB03-2SMB02fSMB03SMB02lSMB03?p=93?p=46

If this is the equation, the limit does not exist, the answer should be DNE! The way you do it is basically plug in 1 and then you will get 0 on the top and basically it won't exist.

 

[BloodyFrozen]

http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:1,SMB02FSMB03x-1SMB10SMB02RSMB03x+3SMB02rSMB03-2SMB02fSMB03SMB02lSMB03?p=93?p=46

If this is the equation, the limit does not exist, the answer should be DNE! The way you do it is basically plug in 1 and then you will get 0 on the top and basically it won't exist.

This function definitely HAS a limit as x approaches 1. Just because it evaluates as 0/0 does not necessarily mean the limit does not exist. Instead, you can use other methods like multiplying by a conjugate, factoring, L'Hôpital's Rule, etc.

with that being said, you could use L'Hôpital's Rule, but I'd recommend multiplying by the conjugate as sheriff89 said.