Confirm the limit algebraically: lim ((1/2+x)-.5)/x x->0
The limit is 2 but Im not sure how to confirm it.
use L'Hopital's rule
((1/(2+x))-.5)/x - Extra parentheses
Looking that up right now, thanks.
\[\frac{ \frac{ 1 }{ 2+x }-5 }{ x }?\]
yes that's right, sorry it wasn't clear.
Oh, theres no 5. Its 1/2.
i c
wait what's 1/2?
The limit zzr0ck3r made. Instead of -5 its -.5
Rewrite that as: (1/(2+x))/x - (1/2/x)
wait I don't see how that helps
Just looked up a L'Hopital video by Khan. I think I get it now.
L'Hopital's rule is the easiest way for these types of limits, seriously
\[\lim_{x \rightarrow 0}\frac{\frac{1}{2+x}-\frac{1}{2}}{x}\] \[\lim_{x \rightarrow 0}\frac{\frac{2}{2*(2+x)}-\frac{(2+x)}{2*(2+x)}}{x}\]
before i continue i must stress ALGEBRAICALLY
the answer is not 2 we have 2-(2+x)/(2*(2+x))/x = -1/(4+2x) run limit = -1/4
Fudge sorry. That was the limit to another question I did.
(2-(2+x))/(2*(2+x))*1/x simplify -1/(4+2x) run limit -1/4
dont use lhopital if you dont know it
\[\lim_{x \rightarrow 0}\frac{\frac{2-2-x}{2*(2+x)}}{x}\] \[\lim_{x \rightarrow 0}\frac{\frac{-x}{2*(2+x)}}{x}\] \[\lim_{x \rightarrow 0}\frac{-1}{2*(2+x)}\]
oh you guys finished it here.. sorry didn't notice.
:) im to fast:P
pretty easy when you arnt using latex:P
zzr0ck3r - *le challenge accepted*
typing it out in latex is annoying but nice to look at
I already won the challenge.:P
Wow that's fantastic Agent. It looks like a great resource. Thanks everyone for helping, Ill make sure I understand it.
yeah I agrea completeidiot even more so with this problem x/y/c/x:)
Wish I could best answer you all.
lol what ever
didnt I answer it twice before those slowpokes?
im jk I could not care less:P
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