Question 908637
a) f(x) = 2^x and g(x) = e^(x^2)
f'(x) = 2^x * log(2)
g'(x) = 2e^(x^2)x
then lim as x approaches infinity is 2^x * log(2) / (2e^(x^2)x = L
we see that lim as x approaches infinity of g(x) is +infinity which satisfies L'Hopital's Theorem criteria and
lim as x approaches infinity is 2^x * log(2) / (2e^(x^2)x = 1
b) lim as x approaches to 0 (sinx)^lncosx is 1