Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 36

Given to solve ,
lim_(x->oo) e^(x/2)/x
As x thends to oo we get e^(x/2)/x = oo/oo
L'Hopital's Rule says if
lim_(x->a) f(x)/g(x) = 0/0 or (+-oo)/(+-oo) then the limit is: lim_(x->a) f'(x)/g'(x)
so , now evaluating
lim_(x->oo) e^(x/2)/x
upon using the L'Hopital's Rule we get
=lim_(x->oo) ((e^(x/2))')/((x)')
=lim_(x->oo) ((e^(x/2))(1/2))/(1)
=>lim_(x->oo) (e^(x/2))/2
now on x-> oo we get e^(x/2) -> oo
so,
lim_(x->oo) (e^(x/2))/2 = oo