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anonymous
 one year ago
Help with limits for a medal?
anonymous
 one year ago
Help with limits for a medal?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im attaching the file

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use the graph to find the limit and determine the continuity of the function.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about that. My internet is giving me some troubles.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2np the servers here are funny sometimes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just a sec. It's not letting me post them

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Right, this is what I figured. Ok as you walk along the graph from left to right there is an obvious problem at \(x=1\). What value do you approach(y value) as you get closer to \(x=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There's a hole in the graph

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2what value is y when you approach from the left?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2so \(\lim_{x\rightarrow c^{}}f(x)=2\) What about from the right?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Yep so \(\lim_{x\rightarrow 2^{+}}f(x)=0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 2 doesn't equal zero.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2obviously not, thus \(\lim_{x\rightarrow c}f(x)\) does not exist

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was much easier than I expected.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2now, since we cant just fill in one point and make it continuous, i.e. there is no one point that we can add so that we can draw the entire graph without lifting the pencil and thus is does not have a removable discontinuity .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see now. Thank you! :)
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