Limit of a limit?

Calculus Level 4

lim k 1 lim a 1 lim n j = 1 k ( a n k ( a + j n ) n ) = ? \large \lim_{k\to 1} \lim_{a\to 1} \lim_{n\to \infty} \prod_{j=1}^{k}\left ( a^{-nk} \left (a+ \dfrac{j}{n}\right)^{n}\right) = \, ?

k e ke 1 ln a \ln a e e

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1 solution

Let X= lim k 1 lim a 1 lim n ( j = 1 k ( a n k ( a + j n ) n ) ) \large \lim_{k\to 1} \lim_{a\to 1} \lim_{n\to \infty} \left(\prod_{j=1}^{k}\left ( a^{-nk}(a+ \frac{j}{n})^{n}\right) \right)

= lim k 1 lim a 1 lim n ( a n k [ ( a + 1 n ) ( a + 2 n ) . . . . ( a + k n ) ] n ) \large \lim_{k\to 1} \lim_{a\to 1} \lim_{n\to \infty} \left(a^{-nk}[(a+ \frac{1}{n})(a+ \frac{2}{n})....(a+ \frac{k}{n})]^{n}\right) = lim k 1 lim a 1 lim n ( [ ( 1 + 1 a n ) n ( 1 + 2 a n ) n . . . . ( 1 + k a n ) n ) \large \lim_{k\to 1} \lim_{a\to 1} \lim_{n\to \infty} \left([(1+ \frac{1}{an})^{n}(1+ \frac{2}{an})^{n}....(1 + \frac{k}{an})^{n}\right) [Introducing each 'a' into all the k brackets]

Now, Taking Log of both sides,

logX = lim k 1 lim a 1 lim n ( n [ l o g ( 1 + 1 a n ) + l o g ( 1 + 2 a n ) . . . . ( 1 + k a n ) ] ) \large \lim_{k\to 1} \lim_{a\to 1} \lim_{n\to \infty} \left(n[log(1+ \frac{1}{an})+log(1+ \frac{2}{an})....(1+ \frac{k}{an})]\right)

Changing n= 1 s \frac{1}{s} , we get

= lim k 1 lim a 1 lim s 0 ( [ l o g ( 1 + s a ) + l o g ( 1 + 2 s a ) . . . . ( 1 + k s a ) ] s ) \large \lim_{k\to 1} \lim_{a\to 1} \lim_{s\to 0} \left(\frac{[log(1+ \frac{s}{a})+log(1+ \frac{2s}{a})....(1+ \frac{ks}{a})]}{s}\right) = lim k 1 lim a 1 lim s 0 ( 1 a [ l o g ( 1 + s a ) s a + 2 a [ l o g ( 1 + 2 s a ) 2 s a . . . . k s a [ l o g ( 1 + k s a ) k s a ] ) \large \lim_{k\to 1} \lim_{a\to 1} \lim_{s\to 0} \left(\frac{1}{a}\frac{[log(1+ \frac{s}{a})}{\frac{s}{a}}+\frac{2}{a}\frac{[log(1+ \frac{2s}{a})}{\frac{2s}{a}}....\frac{ks}{a}\frac{[log(1+ \frac{ks}{a})}{\frac{ks}{a}}]\right)
Since, lim s 0 ( l o g ( 1 + s ) s = 1 ) \lim_{s\to 0} \left(\frac{log(1+s)}{s} = 1\right)

= lim k 1 lim a 1 ( 1 a + 2 a + . . . . . . k a ) ) \large \lim_{k\to 1} \lim_{a\to 1} \left(\frac{1}{a}+\frac{2}{a}+......\frac{k}{a})\right) Therefore, X = lim k 1 lim a 1 ( e k ( k + 1 ) 2 a ) \large \lim_{k\to 1} \lim_{a\to 1} \left(e^{\frac{k(k+1)}{2a}}\right) = e

Nice solution

Jun Arro Estrella - 5 years, 3 months ago

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Oh Thank you !

Aditya Narayan Sharma - 5 years, 3 months ago

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