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*logx-x^2/4].
Applying limits, Def. Int. =(1/2log1-1/4) – [lim as x->0 of(x^2/2*logx-0)] = -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^2-2x-2)].
Applying limits, Def. Int.= lim as x->infinity, of [(-x^2-2x-2)/ e^x] – [lim as x->0 of [ (-x^2-2x-2) /e^x *] = 2, which is the required answer, if I take limit of [1/e^x(-x^2-2x-2]=0 as x->infinity ……(B).
Is it possible to deduce these limits without L’Hopital’s Rule, Squeeze Theorem etc. ?
3 answers
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.
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
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e^-x -> 0 as x->∞, so the upper limit -> 0
e^0 = 1, so the lower limit -> 2
Since x->0 and logx->-∞, x/logx -> 0/-∞ = 0
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e^-x -> 0 as x->∞, so the upper limit -> 0
e^0 = 1, so the lower limit -> 2
Thank you very much, Mr. Steve. It was actually x^2/2*logx but I feel it won't matter.
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