Creative Question II

Calculus Level 2

lim x 201 1 x + 201 2 x + 201 3 x + 201 4 x 201 5 x + 201 6 x + 201 7 x + 201 8 x x = a b \large \lim_{x\to\infty} \sqrt[x]{\frac {2011^x+2012^x+2013^x+2014^x}{2015^x + 2016^x +2017^x+2018^x}} = \frac{a}{b}

Given that a a and b b are positive coprime integers, find the value of a + b a+b .


The answer is 2016.

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1 solution

Chew-Seong Cheong
Jul 31, 2018

L = lim x 201 1 x + 201 2 x + 201 3 x + 201 4 x 201 5 x + 201 6 x + 201 7 x + 201 8 x x = lim x 201 4 x ( ( 2011 2014 ) x + ( 2012 2014 ) x + ( 2013 2014 ) x + 1 ) 201 8 x ( ( 2015 2018 ) x + ( 2016 2018 ) x + ( 2017 2018 ) x + 1 ) x = 201 4 x ( 0 + 0 + 0 + 1 ) 201 8 x ( 0 + 0 + 0 + 1 ) x = 1007 1009 \begin{aligned} L & = \lim_{x \to \infty} \sqrt[x]{\frac {2011^x+2012^x+2013^x+2014^x}{2015^x+2016^x+2017^x+2018^x}} \\ & = \lim_{x \to \infty} \sqrt[x]{\frac {2014^x \left(\left(\frac {2011}{2014}\right)^x+\left(\frac {2012}{2014}\right)^x+\left(\frac {2013}{2014}\right)^x+1\right)}{2018^x \left(\left(\frac {2015}{2018}\right)^x+\left(\frac {2016}{2018}\right)^x+\left(\frac {2017}{2018}\right)^x+1\right)}} \\ & = \sqrt[x]{\frac {2014^x \left(0+0+0+1\right)}{2018^x \left(0+0+0+1\right)}} \\ & = \frac {1007}{1009} \end{aligned}

Therefore, a + b = 1007 + 1009 = 2016 a+b=1007+1009 = \boxed{2016} .

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