This is the question I'm trying to answer
This is how I went about proving i) and ii):
i) limx→∞lnxxr=limx→∞(1xr−1⋅lnxx)=limx→∞(1x−1⋅1xr⋅lnxx)=limx→∞x⋅limx→∞1xr⋅limx→∞lnxx=∞⋅1∞⋅0=1⋅0=0
ii) limx→0+xrlnx=limx→0+(xr−1⋅xlnx)=limx→0+xr−1⋅limx→0+xlnx=limx→∞1xr−1⋅limx→0+xlnx=limx→∞1x−1⋅limx→∞1xr⋅limx→0+xlnx=limx→∞x⋅limx→∞1xr⋅limx→0+xlnx=∞⋅1∞⋅0=1⋅0=0
For i) I wasn't sure if writing limx→∞x as ∞ and limx→∞1xr as 1∞ so that they cancelled out to give 1 was the right thing to do since the limits for x and xr are different as x gets very large, although they both tend to infinity.
For ii) I wasn't sure if writing limx→0+xr−1 as limx→∞1xr−1 was right to do even though they both give the value 0.
Answer
Your question has already given a big hint. You may assume that these results hold for r=1. Thus we are given the following limits limx→∞logxx=0=limx→0+xlogx With these as given the problem is too simple. We have limx→∞logxxr=1rlimx→∞logxrxr=1rlimt→∞logtt=1r⋅0=0 We have used the substitution t=xr and it is given that t→∞ as x→∞. Similarly the second limit is evaluated.
No comments:
Post a Comment