Posted by MS on Thursday, March 13, 2014 at 2:45am.
Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc.
1. Int. of (xlogx)dx from 0 to 1.
Indefinite Int. comes to [x^2/2*logxx^2/4].
Applying limits, Def. Int. =(1/2log11/4) – [lim as x>0 of(x^2/2*logx0)] = 1/4, which is the required answer, if I take limit of (x^2/2*logx)=0 as x>0 ……(A).
2. Int. of (x^2*e^x)dx from 0 to infinity.
Indefinite Int. comes to [e^x(x^22x2)].
Applying limits, Def. Int.= lim as x>infinity, of [(x^22x2)/ e^x] – [lim as x>0 of [ (x^22x2) /e^x *] = 2, which is the required answer, if I take limit of [1/e^x(x^22x2]=0 as x>infinity ……(B).
Is it possible to deduce these limits without L’Hopital’s Rule, Squeeze Theorem etc. ?

I give up  Damon, Thursday, March 13, 2014 at 5:47am
I tried this with algebra a day or two ago and was unable to find the lower limit. I suspect it is zero but was unable to prove it with algebra.

Calculus  Steve, Thursday, March 13, 2014 at 12:45pm
well, x^2/(2logx) = x^2/log(x^2), so if you can figure out the limit of x/logx, you're ok.
Since x>0 and logx>∞, x/logx > 0/∞ = 0

e^x > 0 as x>∞, so the upper limit > 0
e^0 = 1, so the lower limit > 2

Calculus  MS, Friday, March 14, 2014 at 12:14am
Thank you very much, Mr. Steve. It was actually x^2/2*logx but I feel it won't matter.
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