Find limn→∞(n+1)√n+1n√n
First I tried by taking lnyn=ln(n+1)√n+1n√n=√n+1ln(n+1)−√nln(n),
which dose not seems to take me anywhere. Then I tried to use squeeze theorem, since (n+1)√n+1n√n≥1, but I need an upper bound now. It's for a while I am trying to come up with it but stuck. Can you help me please.
Answer
Notice that for f(x)=√xlnx you have
f′(x)=lnx2√x−1√x.
Now by mean value theorem
f(n+1)−f(n)=f′(θn)
for some θn such that $n<\theta_n
If we notice that limx→∞f′(x)=0 we get that
limn→∞(√n+1ln(n+1)−√nlnn)=limn→∞f′(θn)=0.
No comments:
Post a Comment