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. ?
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
Thank you very much, Mr. Steve. It was actually x^2/2*logx but I feel it won't matter.
Yes, it is indeed possible to evaluate these limits without using L'Hopital's Rule or the Squeeze Theorem. In both cases, you can use simple algebraic manipulation and properties of limits to find the answer.
1. Integral of (xlogx)dx from 0 to 1:
To evaluate this integral, you have correctly found the indefinite integral as [x^2/2*logx-x^2/4]. Now, to find the definite integral, you need to substitute the limits of integration.
When plugging in the upper limit, you get [1/2*log(1) - 1/4]. Remember that log(1) is equal to 0, so the first term becomes 0. Simplifying further, you have -1/4.
Now, let's consider the limit (A) as x approaches 0: lim(x^2/2*logx) as x approaches 0. This limit can be evaluated by noticing that as x approaches 0, logx approaches negative infinity. And since x^2 goes to 0 as well, the numerator approaches 0 and the entire limit becomes 0.
Therefore, the final answer for the definite integral is -1/4, which matches the previous result.
2. Integral of (x^2*e^-x)dx from 0 to infinity:
Again, you have correctly found the indefinite integral as [e^-x(-x^2-2x-2)]. To evaluate the definite integral, you need to substitute the limits of integration.
When plugging in the upper limit, you get [(-infinity^2-2*infinity-2)/ e^infinity]. Here, we can observe that as x approaches infinity, both the numerator (-x^2-2x-2) and the denominator (e^x) go to infinity. Therefore, this limit is indeterminate.
For the lower limit, plugging in 0 gives [(-0^2-2*0-2)/ e^0] = -2.
Therefore, to evaluate the definite integral, you need to find the limits (B) as x approaches infinity: lim([1/e^x(-x^2-2x-2)] as x approaches infinity. By applying similar reasoning as before, you can see that both the numerator and denominator approach infinity as x approaches infinity. Hence, this limit is also indeterminate.
In such situations, you may need more advanced techniques like L'Hopital's Rule or the Squeeze Theorem to evaluate these limits precisely. However, if the book has not covered these methods yet, it's possible that it expects you to use these intuitive arguments and recognize that the limit is indeterminate rather than finding the exact value. In this case, indicating that the limit is indeterminate is also an acceptable answer.
Therefore, the final answer for the definite integral is "indeterminate" or "does not exist", given that the limits (B) as x approaches infinity cannot be evaluated without additional techniques.