Limiting Fractional Infinities...

Calculus Level 4

Evaluate:

lim x x x + x 3 x + x 3 x + x 3 \Large \lim_{x \to \infty} \frac { x }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ \cdots } } } }

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4 Complex Solution. 3 1 2 0

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3 solutions

Let l i m x ( x x + x 3 x + x 3 x + x 3 . . . . . . . . . . ) = x x + 1 x 2 3 . x x + x 3 . . . . . . . . . \underset { x\rightarrow \infty }{ lim } \left( \frac { x }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ ..........\infty } } } } \right) \quad = \quad \frac { x }{ x+\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } .\frac { x }{ x\quad +\quad \sqrt [ 3 ]{ x } .........\infty } }

y = x x + 1 x 2 3 . ( y ) \therefore \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { x }{ x+\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } .(y) }

y = x 5 3 x 5 3 + y \therefore \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { { x }^{ \frac { 5 }{ 3 } } }{ { x }^{ \frac { 5 }{ 3 } }\quad +\quad y }

y 2 + ( x 5 3 ) y x 5 3 = 0 \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad { y }^{ 2 }\quad +\quad \left( { x }^{ \frac { 5 }{ 3 } } \right) y\quad -\quad { x }^{ \frac { 5 }{ 3 } }\quad =\quad 0

y = ( x 5 3 ) ± x 10 3 + 4. x 5 3 2 \Rightarrow \quad \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { -\left( { x }^{ \frac { 5 }{ 3 } } \right) \quad \pm \quad \sqrt { { x }^{ \frac { 10 }{ 3 } }\quad +\quad 4.{ x }^{ \frac { 5 }{ 3 } } } }{ 2 }

y = x 5 3 + x 10 3 + 4. x 5 3 2 ( y > 0 ) \Rightarrow \quad \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { -{ x }^{ \frac { 5 }{ 3 } }\quad +\quad \sqrt { { x }^{ \frac { 10 }{ 3 } }\quad +\quad 4.{ x }^{ \frac { 5 }{ 3 } } } }{ 2 } \quad \quad \quad \quad \quad \quad \quad \left( \because \quad y>0 \right)

y = 4 . x 5 3 2 ( x 5 3 + x 10 3 + 4. x 5 3 ) ( R a t i o n a l i s i n g b y d e n o m i n a t o r ) \Rightarrow \quad \quad \quad \quad \quad \quad \quad y\quad =\frac { 4{ .x }^{ \frac { 5 }{ 3 } } }{ 2\left( { x }^{ \frac { 5 }{ 3 } }\quad +\quad \sqrt { { x }^{ \frac { 10 }{ 3 } }\quad +\quad 4.{ x }^{ \frac { 5 }{ 3 } } } \right) } \quad \quad \left( Rationalising\quad by\quad denominator \right)

l i m x y = l i m x 2 { 1 + 1 + 4 x 5 3 } \therefore \quad \quad \quad \quad \quad \quad \quad \quad \quad \underset { x\rightarrow \infty }{ lim } y\quad =\quad \underset { x\rightarrow \infty }{ lim } \frac { 2 }{ \left\{ 1+\sqrt { 1+\frac { 4 }{ { x }^{ \frac { 5 }{ 3 } } } } \right\} }

= 2 1 + 1 =\quad \frac{2}{1 \quad + \quad 1}

l i m x ( x x + x 3 x + x 3 . . . . . . . . . . . . ) = 1 \boxed{\underset { x\rightarrow \infty }{ lim } \left( \frac { x }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\sqrt [ 3 ]{ x } ............\infty } } \right) \quad = \quad 1}

Sandeep Rathod
Feb 13, 2015

l i m x ( x x + x 3 x + x 3 x + x 3 . . . . . . . . . . ) \underset { x\rightarrow \infty }{ lim } \left( \frac { x }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ ..........\infty } } } } \right)

l i m x ( 1 1 + 1 ( x 1 / 3 ) x + x 3 x + x 3 . . . . . . . . . . ) \underset { x\rightarrow \infty }{ lim } \left( \frac { 1 }{ 1+\frac {1}{(x^{1/3} ) x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ ..........\infty } } } } \right)

Thus when x tends to infinity the denominator that is (1 /the cube root equation you can see) will tend to zero

therefore when x tends to infinity the fraction tends to one

Chew-Seong Cheong
Aug 27, 2019

Let y = x + x 3 x + x 3 x + x 3 y = x + \dfrac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{\cdots}}} . Then we have:

y = x + x 3 y y 2 x y x 3 = 0 y = x + x 2 + 4 x 3 2 \begin{aligned} y & = x + \frac {\sqrt[3]x}y \\ y^2 - xy - \sqrt[3]x & = 0 \\ \implies y & = \frac {x+\sqrt{x^2+4\sqrt[3]x}}2 \end{aligned}

Therefore,

L = lim x x x + x 3 x + x 3 x + x 3 = lim x x y = lim x 2 x x + x 2 + 4 x 3 blue Divide up and down by x = lim x 2 1 + 1 + 4 x 5 3 = 1 \begin{aligned} L & = \lim_{x \to \infty} \frac x{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{\cdots}}}} \\ & = \lim_{x \to \infty} \frac xy \\ & = \lim_{x \to \infty} \frac {2x}{x+\sqrt{x^2+4\sqrt[3]x}} & \small \color{#3D99F6} \text{blue} \small \color{#3D99F6} \text{Divide up and down by }x \\ & = \lim_{x \to \infty} \frac 2{1+\sqrt{1+4x^{-\frac 53}}} \\ & = \boxed 1 \end{aligned}

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