Posted by Z32 on Tuesday, September 29, 2009 at 1:43pm.
for the first
as x gets larger, 6/e^x gets closer to zero
so the limit is 0+7 = 7
for the second, remember that 1/a^-n = a^n
so for huge negative values of x, 6/e^x becomes "huge"
so for the second, it is undefined (infinitely large)
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