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Mathematics 15 Online
OpenStudy (anonymous):

Confirm the limit algebraically: lim ((1/2+x)-.5)/x x->0

OpenStudy (anonymous):

The limit is 2 but Im not sure how to confirm it.

OpenStudy (agent47):

use L'Hopital's rule

OpenStudy (anonymous):

((1/(2+x))-.5)/x - Extra parentheses

OpenStudy (anonymous):

Looking that up right now, thanks.

OpenStudy (zzr0ck3r):

\[\frac{ \frac{ 1 }{ 2+x }-5 }{ x }?\]

OpenStudy (anonymous):

yes that's right, sorry it wasn't clear.

OpenStudy (anonymous):

Oh, theres no 5. Its 1/2.

OpenStudy (zzr0ck3r):

i c

OpenStudy (agent47):

wait what's 1/2?

OpenStudy (anonymous):

The limit zzr0ck3r made. Instead of -5 its -.5

OpenStudy (agent47):

Rewrite that as: (1/(2+x))/x - (1/2/x)

OpenStudy (agent47):

wait I don't see how that helps

OpenStudy (anonymous):

Just looked up a L'Hopital video by Khan. I think I get it now.

OpenStudy (agent47):

L'Hopital's rule is the easiest way for these types of limits, seriously

OpenStudy (anonymous):

\[\lim_{x \rightarrow 0}\frac{\frac{1}{2+x}-\frac{1}{2}}{x}\] \[\lim_{x \rightarrow 0}\frac{\frac{2}{2*(2+x)}-\frac{(2+x)}{2*(2+x)}}{x}\]

OpenStudy (anonymous):

before i continue i must stress ALGEBRAICALLY

OpenStudy (zzr0ck3r):

the answer is not 2 we have 2-(2+x)/(2*(2+x))/x = -1/(4+2x) run limit = -1/4

OpenStudy (anonymous):

Fudge sorry. That was the limit to another question I did.

OpenStudy (zzr0ck3r):

(2-(2+x))/(2*(2+x))*1/x simplify -1/(4+2x) run limit -1/4

OpenStudy (agent47):

http://www.twiddla.com/929034

OpenStudy (zzr0ck3r):

dont use lhopital if you dont know it

OpenStudy (anonymous):

\[\lim_{x \rightarrow 0}\frac{\frac{2-2-x}{2*(2+x)}}{x}\] \[\lim_{x \rightarrow 0}\frac{\frac{-x}{2*(2+x)}}{x}\] \[\lim_{x \rightarrow 0}\frac{-1}{2*(2+x)}\]

OpenStudy (agent47):

oh you guys finished it here.. sorry didn't notice.

OpenStudy (zzr0ck3r):

:) im to fast:P

OpenStudy (zzr0ck3r):

pretty easy when you arnt using latex:P

OpenStudy (agent47):

zzr0ck3r - *le challenge accepted*

OpenStudy (anonymous):

typing it out in latex is annoying but nice to look at

OpenStudy (zzr0ck3r):

I already won the challenge.:P

OpenStudy (anonymous):

Wow that's fantastic Agent. It looks like a great resource. Thanks everyone for helping, Ill make sure I understand it.

OpenStudy (zzr0ck3r):

yeah I agrea completeidiot even more so with this problem x/y/c/x:)

OpenStudy (anonymous):

Wish I could best answer you all.

OpenStudy (zzr0ck3r):

lol what ever

OpenStudy (zzr0ck3r):

didnt I answer it twice before those slowpokes?

OpenStudy (zzr0ck3r):

im jk I could not care less:P

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