is the discontinuity of (e^(2x) -1) / (e^x - 1) at x= 0 removable? if not explain why not, and if yes, give a value for f(0) that will make f(x) continuous
ok so i'm confused, so the limits as x approaches 0 from either side has to be the same? or something like that to remove x=0 ?
ok so i'm confused, so the limits as x approaches 0 from either side has to be the same? or something like that to remove x=0 ?
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The numerator can factor as (e^x - 1)*(e^x + 1). Thus, we have
f(x) = (e^x - 1)*(e^x + 1) / (e^x - 1) = e^x + 1. So f(0) = e^0 + 1 = 2.
Set f(0) = 2 to make f(x) continuous.
f(x) = (e^x - 1)*(e^x + 1) / (e^x - 1) = e^x + 1. So f(0) = e^0 + 1 = 2.
Set f(0) = 2 to make f(x) continuous.
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Yes, it is.
Notice that the numerator factors as (e^x+1)(e^x-1). Now, you can remove the common factor (e^x-1).
Then, define f(0) = e^0+1 = 1+1 =2
Notice that the numerator factors as (e^x+1)(e^x-1). Now, you can remove the common factor (e^x-1).
Then, define f(0) = e^0+1 = 1+1 =2