Crazy Limits!!! 1

Calculus Level 5

Let

I = lim x 0 ( 1 x 5 0 x e t 2 d t 1 x 4 + 1 3 x 2 ) \displaystyle I = \lim_{x \to 0}\left(\frac{1}{x^{5}} \int_0^{x} e^{-t^{2}}dt -\frac{1}{x^{4}} + \frac{1}{3x^{2}} \right)

and

K = lim n ( [ 1 2 7 ] + [ 2 2 7 ] + + [ n 2 7 ] n 3 ) . \displaystyle K= \lim_{n \to \infty}\left(\frac{[1^{2}\sqrt{7}] + [2^{2}\sqrt{7}]+ \ldots + [n^{2}\sqrt{7}]}{n^{3}} \right) .

Then find

π 2 × I + 9 K 2 4 . \displaystyle \pi^{2} \times I + \frac{9K^{2}}{4}.

Details and Assumptions

  • [ x ] [x] denotes the floor function.

  • Use π 2 = 10 \pi^2 = 10 .


The answer is 2.75.

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1 solution

Ayush Verma
Oct 18, 2014

Actually,this is not as hard as this solution seems.So at least have a look e t 2 = r = 0 ( t 2 ) r r ! = r = 0 ( 1 ) r t 2 r r ! 0 x e t 2 d t = [ r = 0 ( 1 ) r 2 r + 1 t 2 r + 1 r ! ] 0 x = r = 0 ( 1 ) r 2 r + 1 x 2 r + 1 r ! = r = 0 C r x 2 r + 1 w h e r e C r = ( 1 ) r 2 r + 1 1 r ! I = l i m x 0 0 x e t 2 d t x + 1 3 x 3 x 5 = l i m x 0 ( x 1 3 x 3 + 1 10 x 5 + r = 3 C r x 2 r + 1 ) x + 1 3 x 3 x 5 I = l i m x 0 ( 1 10 + C 3 x 2 + C 4 x 4 + C 5 x 6 + . . . . ) = 1 10 f o r r > 0 r 2 7 1 [ r 2 7 ] r 2 7 + 1 r = 1 n ( r 2 7 1 ) r = 1 n [ r 2 7 ] r = 1 n ( r 2 7 + 1 ) 2 n 3 + 3 n 2 + n 6 7 n r = 1 n [ r 2 7 ] 2 n 3 + 3 n 2 + n 6 7 + n l i m n ( 2 n 3 + 3 n 2 + n 6 7 n ) n 3 l i m n r = 1 n [ r 2 7 ] n 3 l i m n ( 2 n 3 + 3 n 2 + n 6 7 + n ) n 3 7 3 K 7 3 K = 7 3 A n s = π 2 I + 9 4 K 2 = 10 × 1 10 + 9 4 × 7 9 = 2.75 \large { e }^{ -{ t }^{ 2 } }=\sum _{ r=0 }^{ \infty } \cfrac { { ({ -t }^{ 2 }) }^{ r } }{ r! } =\sum _{ r=0 }^{ \infty }{ { (-1) }^{ r } } \cfrac { { t }^{ 2r } }{ r! } \\ \\ \Rightarrow \int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 } }dt= } \left[ \sum _{ r=0 }^{ \infty }{ \cfrac { { (-1) }^{ r } }{ 2r+1 } } \cfrac { { t }^{ 2r+1 } }{ r! } \right] _{ 0 }^{ x }=\sum _{ r=0 }^{ \infty }{ \cfrac { { (-1) }^{ r } }{ 2r+1 } } \cfrac { { x }^{ 2r+1 } }{ r! } \\ \\ \quad \quad \quad \quad \quad \quad \quad \quad =\sum _{ r=0 }^{ \infty }{ { C }_{ r } } { x }^{ 2r+1 }\quad \quad \quad where\quad { C }_{ r }=\cfrac { { (-1) }^{ r } }{ 2r+1 } \cfrac { 1 }{ r! } \\ \\ \Rightarrow I=\underset { x\rightarrow 0 }{ lim } \cfrac { \int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 } }dt-x+\cfrac { 1 }{ 3 } { x }^{ 3 } } }{ { x }^{ 5 } } =\underset { x\rightarrow 0 }{ lim } \cfrac { \left( x-\cfrac { 1 }{ 3 } { x }^{ 3 }+{ \cfrac { 1 }{ 10 } x }^{ 5 }+\sum _{ r=3 }^{ \infty }{ { C }_{ r } } { x }^{ 2r+1 } \right) -x+\cfrac { 1 }{ 3 } { x }^{ 3 } }{ { x }^{ 5 } } \\ \\ \quad \quad \quad I=\underset { x\rightarrow 0 }{ lim } \left( \cfrac { 1 }{ 10 } +{ C }_{ 3 }{ x }^{ 2 }{ +C }_{ 4 }{ x }^{ 4 }+{ C }_{ 5 }{ x }^{ 6 }+.... \right) =\cfrac { 1 }{ 10 } \\ \\ for\quad r>0\\ \\ { r }^{ 2 }\sqrt { 7 } -1\le \left[ { r }^{ 2 }\sqrt { 7 } \right] \le { r }^{ 2 }\sqrt { 7 } +1\\ \\ \Rightarrow \sum _{ r=1 }^{ n }{ ({ r }^{ 2 }\sqrt { 7 } -1) } \le \sum _{ r=1 }^{ n }{ \left[ { r }^{ 2 }\sqrt { 7 } \right] } \le \sum _{ r=1 }^{ n }{ ({ r }^{ 2 }\sqrt { 7 } +1) } \\ \\ \Rightarrow \cfrac { 2{ n }^{ 3 }+3{ n }^{ 2 }+n }{ 6 } \sqrt { 7 } -n\quad \le \quad \sum _{ r=1 }^{ n }{ \left[ { r }^{ 2 }\sqrt { 7 } \right] } \quad \le \quad \cfrac { 2{ n }^{ 3 }+3{ n }^{ 2 }+n }{ 6 } \sqrt { 7 } +n\\ \\ \Rightarrow \underset { n\rightarrow \infty }{ lim } \cfrac { \left( \cfrac { 2{ n }^{ 3 }+3{ n }^{ 2 }+n }{ 6 } \sqrt { 7 } -n \right) }{ n^{ 3 } } \le \underset { n\rightarrow \infty }{ lim } \cfrac { \sum _{ r=1 }^{ n }{ \left[ { r }^{ 2 }\sqrt { 7 } \right] } }{ n^{ 3 } } \le \underset { n\rightarrow \infty }{ lim } \cfrac { \left( \cfrac { 2{ n }^{ 3 }+3{ n }^{ 2 }+n }{ 6 } \sqrt { 7 } +n \right) }{ n^{ 3 } } \\ \\ \Rightarrow \cfrac { \sqrt { 7 } }{ 3 } \le K\le \cfrac { \sqrt { 7 } }{ 3 } \quad \Rightarrow K=\cfrac { \sqrt { 7 } }{ 3 } \\ \\ Ans={ \pi }^{ 2 }I+\cfrac { 9 }{ 4 } { K }^{ 2 }=10\times \cfrac { 1 }{ 10 } +\cfrac { 9 }{ 4 } \times \cfrac { 7 }{ 9 } =2.75

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