What is #lim_(x->oo) e^x/(x-1)# ?
2 Answers
Explanation:
If we evaluate this by direct substitution we get
This answer is not helpful. Since
L'Hospital's rule provides a solution for us. L'Hospital's rule states that:
In other words, if and only if a limit ends up as
Applying the rule,
Explanation:
The simple answer is that
Hence:
#lim_(x->oo) e^x/(x-1) = oo#
A little more formally:
#e^x = sum_(k=0)^oo x^k/(k!) > x^2/2#
Hence:
#lim_(x->oo) e^x/(x-1) >= lim_(x->oo) (x^2/2)/(x-1) >= lim_(x->oo) 1/2(x^2/x) = lim_(x->oo) x/2 = oo#