Limited yet powerful (VIII)

Calculus Level 4

lim x 0 ( e x 2 / 2 cos x ) 1 / x 4 \large \lim_{x\to 0} (e^{x^2 /2} \cos x)^{1/x^4}

If the value of the limit above is equal to e a / b e^{-a/b} , where a a and b b are coprime positive integers, find the value of a + b a+b .


The answer is 13.

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2 solutions

Rishi Sharma
Mar 6, 2016

lim x > 0 ( e x 2 2 c o s x ) 1 x 4 = e lim x > 0 e x 2 2 c o s x 1 x 4 n o w u s i n g e x = n = 0 x n ! n a n d c o s x = n = 0 ( 1 ) n x 2 n ( 2 n ) ! t h e l i m i t t r a n s f o r m s t o e lim x > 0 x 4 12 + . . . . x 4 = e 1 12 s o a = 1 a n d b = 12 h e n c e a + b = 13 \lim _{ x->0 }{ { \left( { e }^{ { \frac { x }{ 2 } }^{ 2 } }cosx \right) }^{ \frac { 1 }{ { x }^{ 4 } } } } ={ e }^{ \lim _{ x->0 }{ { \frac { { e }^{ \frac { { x }^{ 2 } }{ 2 } }cosx-1 }{ { x }^{ 4 } } } } }\\ now\quad using\\ { e }^{ x }=\sum _{ n=0 }^{ \infty }{ { \frac { x }{ n! } }^{ n } } \quad and\quad cosx=\sum _{ n=0 }^{ \infty }{ { { \left( -1 \right) }^{ n }\frac { { x }^{ 2n } }{ (2n)! } } } \\ the\quad limit\quad transforms\quad to\quad \\ { e }^{ \lim _{ x->0 }{ \frac { \frac { { -x }^{ 4 } }{ 12 } +.... }{ { x }^{ 4 } } } }={ e }^{ \frac { -1 }{ 12 } }\quad so\quad a=1\quad and\quad b=12\\ hence\quad a+b=13

Noel Lo
Dec 27, 2015

See below for the step-by-step solution:

Done it same way and used series of tanx at step 7..

rishabh singhal - 4 years, 10 months ago

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