I need to know how to calculate this without using l'hospitals rule:
limit as x approaches infinity of: x(a1/x−1)
I saw that the answer is log(a), but I want to know how they got it.
The book implies that I should be able to find it by just using algebraic manipulation and substitution.
Answer
METHOD 1:
limx→∞x(a1/x−1)=limx→0+ax−1x=limx→0+exloga−1x=limx→0+(1+(loga)x+O(x2))−1x=limx→0+(loga+O(x))=loga
METHOD 2:
Another way to do this is to substitute y=ax in (1). Then
limx→∞x(a1/x−1)=limx→0+ax−1x=limy→1+y−1logy/loga=logalimy→1+y−1logy
Noting that for y>1, y−1y≤logy≤y−1. Then,
1≤y−1logy≤y
and the squeeze theorem does the rest!
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