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Mathematics 18 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

@xapproachesinfinity

OpenStudy (anonymous):

@Loser66

OpenStudy (anonymous):

@caozeyuan

OpenStudy (xapproachesinfinity):

old man knows the answer :)

OpenStudy (xapproachesinfinity):

hehehe, i will try ;(

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.

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!

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 :)

OpenStudy (xapproachesinfinity):

it just about understanding the concepts, definitions

OpenStudy (xapproachesinfinity):

good luck :)

OpenStudy (anonymous):

Thanks again! :)

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