If f is positive and differentiable in (0,∞), then I want to find the following limit.
limn→∞(f(x+1n)f(x))n.
I have done as follows:
limn→∞(f(x+1n)f(x))n=(limn→∞f(x+1n)f(x))n=1 as f is continuous at x=1n. Am I right? I doubt. Please help!
Answer
log[(f(x+1n)f(x))n]=nlogf(x)+1nf′(x)+ϵ(1n)1nf(x)=nlog[1+1nf′(x)f(x)+ϵ(1n)1n]
with lim0ϵ=0.
∼n1nf′(x)f(x)=f′(x)f(x)
so the limit is expf′(x)f(x)
No comments:
Post a Comment