Basic limit

Calculus Level 3

lim ⁡ x → 1 4 x 3 + 3 x + 2 − 3 x 2 + x + 5 x 2 + x − 2 = a b \large \lim _{ x \to 1}{ \frac { \sqrt { 4{ x }^{ 3 }+3x+2 } -\sqrt { 3{ x }^{ 2 }+x+5 } }{ { x }^{ 2 }+x-2 } } = \frac { a }{ b }

If the equation above holds true for coprime positive integers a a and b b , what is a + b a + b ?


The answer is 13.

Alvin Willio by Alvin Willio , 25, Indonesia

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2 solutions

Chew-Seong Cheong
May 16, 2017

L = lim ⁡ x → 1 4 x 3 + 3 x + 2 − 3 x 2 + x + 5 x 2 + x − 2 = lim ⁡ x → 1 ( 4 x 3 + 3 x + 2 − 3 x 2 + x + 5 ) ( 4 x 3 + 3 x + 2 + 3 x 2 + x + 5 ) ( x 2 + x − 2 ) ( 4 x 3 + 3 x + 2 + 3 x 2 + x + 5 ) = lim ⁡ x → 1 4 x 3 + 3 x + 2 − 3 x 2 − x − 5 ( x 2 + x − 2 ) ( 4 x 3 + 3 x + 2 + 3 x 2 + x + 5 ) = lim ⁡ x → 1 4 x 3 − 3 x 2 + 2 x − 3 ( x 2 + x − 2 ) ( 4 x 3 + 3 x + 2 + 3 x 2 + x + 5 ) = lim ⁡ x → 1 ( x − 1 ) ( 4 x 2 + x + 3 ) ( x − 1 ) ( x + 2 ) ( 4 x 3 + 3 x + 2 + 3 x 2 + x + 5 ) = 8 3 ( 3 + 3 ) = 4 9 \begin{aligned} L & = \lim_{x \to 1} \frac {\sqrt{4x^3+3x+2}-\sqrt{3x^2+x+5}}{x^2+x-2} \\ & = \lim_{x \to 1} \frac {\left(\sqrt{4x^3+3x+2}-\sqrt{3x^2+x+5}\right)\color{#3D99F6}\left(\sqrt{4x^3+3x+2}+\sqrt{3x^2+x+5}\right)}{\left(x^2+x-2 \right) \color{#3D99F6} \left(\sqrt{4x^3+3x+2} +\sqrt{3x^2+x+5}\right)} \\ & = \lim_{x \to 1} \frac {4x^3+3x+2-3x^2-x-5}{\left(x^2+x-2 \right)\left(\sqrt{4x^3+3x+2} +\sqrt{3x^2+x+5}\right)} \\ & = \lim_{x \to 1} \frac {4x^3-3x^2+2x-3}{\left(x^2+x-2 \right)\left(\sqrt{4x^3+3x+2} +\sqrt{3x^2+x+5}\right)} \\ & = \lim_{x \to 1} \frac {\cancel{(x-1)}(4x^2+x+3)}{\cancel{(x-1)}(x+2) \left(\sqrt{4x^3+3x+2} +\sqrt{3x^2+x+5}\right)} \\ & = \frac 8{3(3+3)} = \frac 49 \end{aligned}

⟹ a + b = 4 + 9 = 13 \implies a+b = 4+9 = \boxed{13}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Rudraksh Sisodia
Mar 27, 2016

just use L hospital rule for the first time and then substitute 1 ,,, the anwser seems to be there !!!

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Well I guess that's the simplest of all methods for this particular method.

Rahul Singh - 4 years ago

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