how do i find the limit ln(x-3) x-->3+

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how do i find the limit ln(x-3) x-->3+

Mathematics
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it's approaching from the right, so it's defined because x-3 is positive and what number does lny approach as y approaches zero?
ln 0 = infinity
...because x^a is always positive for x>0 we can only make it small by making 'a' large negative

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Other answers:

oh kinda get it
you migth think of it like this. the log is the inverse of the exponential. when would \[e^x=0\]? well never, but it would be close to zero if x went to minus infinity
yes it should be minus^ Igb is a bit off
so as x goes to minus infinity, \[e^x\] goes to zero and as x goes to zero, \[\ln(x)\] goes to minus infinity

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