How to prove this limit (see details below)?

Question

Suppose f is differentiable on [0, oo) and lim (x --> oo) k f(x) + f'(x) = L, where L is finite and k >0. Then, prove that lim (x --> oo) f(x) = L/k and lim (x --> oo) f'(x) = 0

Answer 1

Well, I answered exactly this same question one hour ago. Anyway, this is my answer:

First, let's consider the case L <>0 (different from 0). Define g(x) = e^(kx) and h(x) = f(x) e^(kx) . Then, g never vanishes, f(x) = h(x)/g(x) and , since k >0, g(x) -> oo as x -> oo. In addition,

g'(x) = k e^(kx) and h'(x) = e^k(x) [k f(x) + f'(x)], so that

h'(x)/g'(x) = (k f(x) + f'(x))/k. and, therefore,

lim (x -> oo) h'(x)/g'(x) = lim (x ->oo)(k f(x) + f'(x))/k = L/k

Since L<> 0, lim (x --> oo) k f(x) + f'(x) = L and e^(kx) -> oo as x -> oo, it follows that h'(x) = e^k(x) [k f(x) + f'(x)] goes to oo, if L>0, and to -oo, if L <0. Therefore, the conditions required by L'Hopital rule are fully satisfied, and the existence of lim (x -> oo) h'(x)/g'(x) implies that

lim (x -> oo) h(x)/g(x) = lim ( x -> oo) f(x) = lim (x-> oo) h'(x)/g'(x) = L/k, so that lim (x -> oo) f(x) = L/k. From this, it follows immediately that lim (x -> oo) f'(x) = L - k lim(x -< oo) f(x) = L - k L/k = L - L = 0, completing the proof for the case L <> 0.

Now, let's show the conclusion remains true even if L = 0.

Choose a M >0 and define w(x) = f(x) + M/k. Then, w'(x) = f'(x)and limit (x -> oo) kw(x) + w'(x) = lim (x -> oo) kf(x) + M + f'(x) = lim (x -> oo) (k f(x) + f'(x)) + M = 0 + M = M. Since M <>0, it follows from our previous conclusion that lim ( x -> oo) w(x) = M/k and lim (x -> oo) w'(x) = lim (x -> oo) f'(x) = 0. And this implies immediately that lim (x -> oo) f(x) = 0, which shows the conclusion remains true if L = 0.