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Mathematics 9 Online
OpenStudy (anonymous):

Help with limits for a medal?

OpenStudy (jhannybean):

No question.

OpenStudy (anonymous):

Im attaching the file

OpenStudy (anonymous):

OpenStudy (zzr0ck3r):

Still no question

OpenStudy (anonymous):

Use the graph to find the limit and determine the continuity of the function.

OpenStudy (anonymous):

Sorry about that. My internet is giving me some troubles.

OpenStudy (zzr0ck3r):

What limit?

OpenStudy (zzr0ck3r):

np the servers here are funny sometimes.

OpenStudy (anonymous):

Just a sec. It's not letting me post them

OpenStudy (anonymous):

Hah! Got it!

OpenStudy (zzr0ck3r):

Right, 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 \(x=-1\)

OpenStudy (anonymous):

There's a hole in the graph

OpenStudy (zzr0ck3r):

what value is y when you approach from the left?

OpenStudy (anonymous):

2

OpenStudy (zzr0ck3r):

so \(\lim_{x\rightarrow c^{-}}f(x)=2\) What about from the right?

OpenStudy (anonymous):

It gets closer to 0

OpenStudy (zzr0ck3r):

Yep so \(\lim_{x\rightarrow 2^{+}}f(x)=0\)

OpenStudy (zzr0ck3r):

now does \(2\ne 0\)?

OpenStudy (anonymous):

Yes, 2 doesn't equal zero.

OpenStudy (zzr0ck3r):

obviously not, thus \(\lim_{x\rightarrow c}f(x)\) does not exist

OpenStudy (anonymous):

That was much easier than I expected.

OpenStudy (zzr0ck3r):

now, 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 .

OpenStudy (anonymous):

I see now. Thank you! :)

OpenStudy (zzr0ck3r):

np

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