Inverted Integrated Limits

Calculus Level 4

Let a n = ∫ 1 n + 1 1 n tan ⁡ − 1 ( n x ) d x \displaystyle { a }_{ n }=\int _{ \frac { 1 }{ n+1 } }^{ \frac { 1 }{ n } }{ \tan ^{ -1 }{ (nx) } } dx and b n = ∫ 1 n + 1 1 n sin ⁡ − 1 ( n x ) d x \displaystyle { b }_{ n }=\int _{ \frac { 1 }{ n+1 } }^{ \frac { 1 }{ n } }{ \sin ^{ -1 }{ (nx) } } dx , then lim ⁡ n → ∞ a n b n \displaystyle \lim _{ n\rightarrow \infty }{ \frac { { a }_{ n } }{ { b }_{ n } } } has the value equal to

-1 1 0 1 2 \frac{1}{2}

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Arpita Karkera by Arpita Karkera , 24, India

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1 solution

Hasan Kassim
Aug 5, 2015

Use the Substitution u = n x u=nx in both integrals:

a n = 1 n ∫ n n + 1 1 tan ⁡ − 1 ( u ) d u \displaystyle a_n = \frac{1}{n} \int_{\frac{n}{n+1}}^{1} \tan ^{-1} (u) du

b n = 1 n ∫ n n + 1 1 sin ⁡ − 1 ( u ) d u \displaystyle b_n = \frac{1}{n} \int_{\frac{n}{n+1}}^{1} \sin ^{-1} (u) du

lim ⁡ n → ∞ a n b n = lim ⁡ n → ∞ 1 n ∫ n n + 1 1 tan ⁡ − 1 ( u ) d u 1 n ∫ n n + 1 1 sin ⁡ − 1 ( u ) d u \displaystyle \lim_{n\to \infty } \frac{a_n}{b_n} = \lim_{n\to \infty } \frac{\frac{1}{n} \int_{\frac{n}{n+1}}^{1} \tan ^{-1} (u) du }{\frac{1}{n} \int_{\frac{n}{n+1}}^{1} \sin ^{-1} (u) du}

= lim ⁡ n → ∞ ∫ n n + 1 1 tan ⁡ − 1 ( u ) d u ∫ n n + 1 1 sin ⁡ − 1 ( u ) d u \displaystyle = \lim_{n\to \infty } \frac{ \int_{\frac{n}{n+1}}^{1} \tan ^{-1} (u) du }{\int_{\frac{n}{n+1}}^{1} \sin ^{-1} (u) du}

Direct Substitution Yields an Indetermined form:

= L’Hopital Rule lim ⁡ n → ∞ d d n ∫ n n + 1 1 tan ⁡ − 1 ( u ) d u d d n ∫ n n + 1 1 sin ⁡ − 1 ( u ) d u \displaystyle \overset{{\color{#D61F06}{\text{L'Hopital Rule}}}}{=} \lim_{n\to \infty } \frac{ \frac{d}{dn} \int_{\frac{n}{n+1}}^{1} \tan ^{-1} (u) du }{\frac{d}{dn} \int_{\frac{n}{n+1}}^{1} \sin ^{-1} (u) du}

= Fundamental Theorem of Calculus lim ⁡ n → ∞ − ( n n + 1 ) ′ tan ⁡ − 1 ( n n + 1 ) − ( n n + 1 ) ′ sin ⁡ − 1 ( n n + 1 ) \displaystyle \overset{{\color{#D61F06}{\text{Fundamental Theorem of Calculus}}}}{=} \lim_{n\to \infty } \frac{ - (\frac{n}{n+1})' \tan ^{-1} (\frac{n}{n+1} ) }{ - (\frac{n}{n+1})' \sin ^{-1} (\frac{n}{n+1} ) }

= lim ⁡ n → ∞ tan ⁡ − 1 ( n n + 1 ) sin ⁡ − 1 ( n n + 1 ) \displaystyle = \lim_{n\to \infty } \frac{\tan ^{-1} (\frac{n}{n+1} )}{\sin ^{-1} (\frac{n}{n+1} )}

= tan ⁡ − 1 ( 1 ) sin ⁡ − 1 ( 1 ) \displaystyle = \frac{\tan^{-1} (1)}{\sin^{-1} (1)}

= 1 2 \displaystyle =\boxed{ \frac{1}{2}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Let's use the fact that for a continuous, integrable function f ( x ) f(x) , lim ⁡ a → 1 1 1 − a ∫ a 1 f ( x ) d x = f ( 1 ) \lim_{a \rightarrow 1 } \frac{1}{1-a} \int_a ^ 1 f(x) \, dx = f(1) .

From the first line, we can conclude that since a n = n + 1 n × ( 1 n + 1 ∫ n n + 1 1 tan ⁡ − 1 ( u ) d u ) a_n = \frac{n+1}{n} \times \left( \frac{ 1}{n+1} \int_{\frac{n}{n+1}}^{1} \tan ^{-1} (u) du \right) , hence lim ⁡ n ( n + 1 ) a n = lim ⁡ n + 1 1 × ∫ n n + 1 1 tan ⁡ − 1 ( u ) d u = tan ⁡ − 1 1 \lim n(n+1) a_n = \lim \frac{n+1} {1} \times \int_{\frac{n}{n+1}}^{1} \tan ^{-1} (u) du =\tan^{-1} 1 .

Similarly, lim ⁡ n ( n + 1 ) b n = sin ⁡ − 1 1 \lim n(n+1) b_n = \sin^{-1} 1 .

Challenge Master: I don't see how you computed the limit of a n a_n alone, it is zero, isn't it? Please I want some clarification.

@Calvin Lin

Hasan Kassim - 5 years, 10 months ago

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Oh sorry, I mixed up the 1 n \frac{1}{n} . It would be lim ⁡ n 2 a n = tan ⁡ − 1 1 \lim n^2 a_n = \tan^{-1} 1 instead. Let me update that.

Calvin Lin Staff - 5 years, 10 months ago

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Thanks I got the point :) this fact is very useful .

Hasan Kassim - 5 years, 10 months ago

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