Suppose lim (as x approaches 0-) f(x)=3 and lim (as x approaches 0+) f(x)=4. What happens in the graph of f when x = 0?
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OpenStudy (anonymous):
@xapproachesinfinity
OpenStudy (anonymous):
@Loser66
OpenStudy (anonymous):
@caozeyuan
OpenStudy (xapproachesinfinity):
old man knows the answer :)
OpenStudy (xapproachesinfinity):
hehehe, i will try ;(
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OpenStudy (anonymous):
Who is old man?? :P
OpenStudy (xapproachesinfinity):
f is discontinued at 0
OpenStudy (xapproachesinfinity):
that's what they want you to realize
OpenStudy (xapproachesinfinity):
do you want me to explain it further?
OpenStudy (loser66):
yes, please.
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OpenStudy (anonymous):
Yes please:)
OpenStudy (xapproachesinfinity):
ok,
we are given \[\lim_{x\to 0^-}f(x)=3~~and~~ \lim_{x\to 0^+}f(x)=4\]
Hence \[\lim_{x\to 0^-}f(x) \ne\lim_{x\to 0^+}f(x)\]
then, \[\lim_{x\to 0}f(x) ~~~does~not~exist\]
it reasonable then to say that f is not continuous at 0
OpenStudy (anonymous):
Ohhhh I get it! Thank you! :)
OpenStudy (xapproachesinfinity):
recall the continuity def:
let f be a function from R to R
f is said to be continuous at x0 if and only if the three condition meet:
1) f(x0) exist
2) \lim f(x) exist as x goes to x0
3) lim f(x)=f(x0)
OpenStudy (xapproachesinfinity):
np!
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OpenStudy (loser66):
Woah!! kid is the best. :)
graphing it, please.
OpenStudy (xapproachesinfinity):
|dw:1451680373230:dw|
suppose this to be our function y=f(x)
OpenStudy (anonymous):
xapproachesinfinity you're the best! :)
OpenStudy (xapproachesinfinity):
you can see the jump, which indicates the discontinuity
OpenStudy (xapproachesinfinity):
thanks :)
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OpenStudy (xapproachesinfinity):
it just about understanding the concepts, definitions