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Give the value of x where f has a removable discontinuity
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x = 1
no, its x = 0
is that because the 1 has a black circle?
yes so u can move that circle to 0 instead
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hence the name
yep x=0
what abt the value of x where there is jump discontinuity? thanks!
that would be 1
?
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should be x = 2
whoops I go that wrong; was looking at the vertical asymptote by mistake
at 1 theres infinite discontinuity
do you always take teh value w the black dot because that is an actual pt?
thanks for your help!
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