Limit of an Integral!
n → ∞ lim n ∫ 1 e x 2 ( ln ( x ) ) n d x
For integer n , let L denote the value of the limit above. Find the value of ln ( L ) .
The answer is 3.
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3 solutions
Excellent solution. I hadn't thought to bound it. I used the same substitution, but made e 3 z a Taylor series. I then distributed the limit and integral across the series since it converges uniformly, leaving me with the series representation for e 3 .
MIND BLOWING MAN....Take a bow wanna give u 5 upvotes sorry I can upvote only once :(!!!!
Moderator note:
The top half is correct, but the latter half shows a very common misconception held by students.
The top half is correct, but the latter half shows a very common misconception held by students.
In particular, the claim of " n → ∞ , I n − 1 → I n is wrong. When we cannot say that
n → ∞ lim I n − 1 = I n
because the RHS is meaningless. What is the actual mathematically correct statement that you want to say?
Note: It need not be true that lim n → ∞ I n − 1 = lim n → ∞ I n . It is only true if the limit exists, which has yet to be shown.
Furthermore, there are many sequences that satisfy I n = 3 e 3 − 3 n I n − 1 . However, I_n = e^3 \frac{n+3} is not one of them.
I took ln(x)=t then I used reduction formula generated a pattern and took out e^(3x) common ,rest came out one...so on taking log answer came 3
Let the integral for a particular value of n be f (n). So solve f (n+1) by integration by parts. So we get a relation in terms of in the terms of f (n)and f (n+1). Now let f (n)=n×g (n) and since n tends to infinty Take g (n) and g (n+1) equal and find its value. It e^3. So the answer follows
Hey can you post your solution involving reduction formula. I tried doing doing that way but didn't succeed.
By change of variables x → exp ( z ) , the integral becomes I n = n ∫ 0 1 z n e 3 z d z Now, we will upper-bound and lower-bound the integral. Since z ≤ 1 throughout the range of the integral, we have, I n ≤ n ∫ 0 1 z n e 3 d z ≤ e 3 ( ∗ ) Similarly, from integration by-parts we obtain I n = n + 1 n e 3 − n + 1 3 n ∫ 0 1 e 3 z z n + 1 d z ≥ n + 1 n e 3 − ( n + 1 ) ( n + 2 ) 3 n e 3 Taking limit as n → ∞ , we have n → ∞ lim I n ≥ e 3 ( ∗ ∗ ) The result now follows from ( ∗ ) and ( ∗ ∗ ) .