Infinity Limitation

Calculus Level 3

lim n ( 8 n + 4 n + 2 n 3 2 n ) = ? \large\lim_{n\to\infty}\left(\sqrt[3]{8^n+4^n+2^n}-2^n\right)=\, ?

Give your answer to 3 decimal places.


The answer is 0.333.

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2 solutions

Rishabh Jain
May 16, 2016

Let the limit be H \mathfrak H . Substitute 2 n = x \star~2^n=x followed by using binomial approximation ( 1 + α ) n = 1 + n α ( as α 0 ) \star~(1+\color{#20A900}{\alpha})^n=1+n\color{#20A900}{\alpha}~~(\text{as }\alpha\rightarrow 0) .

H = lim x ( x 3 + x 2 + x 3 x ) = lim x ( x ( 1 + 1 x + 1 x 2 3 1 ) ) = lim x ( x ( 1 + 1 3 ( 1 x + 1 x 2 ) 1 ) ) = lim x ( 1 3 + 1 3 x 0 ) 0.333 \begin{aligned}\mathfrak{H}=& \lim_{x\to\infty}\left(\sqrt[3]{x^3+x^2+x}-x\right)\\=&\lim_{x\to\infty}\left(x\left(\sqrt[3]{1+\color{#20A900}{\dfrac{1}{x}+\dfrac{1}{x^2}}}-1\right)\right)\\=&\lim_{x\to\infty}\left(x\left(\cancel{1}+\dfrac{1}{3}\left(\color{#20A900}{\dfrac{1}{x}+\dfrac{1}{x^2}}\right)-\cancel{1}\right)\right)\\=&\lim_{x\to\infty}\left(\dfrac{1}{3}+\underbrace{\dfrac{1}{3x}}_0\right)\approx\large\color{#0C6AC7}{0.333}\end{aligned}

Same approach :-)

展豪 張 - 5 years, 1 month ago

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Great.... ¨ \ddot\smile

Rishabh Jain - 5 years, 1 month ago

Nice solution. i used L'Hospital's rule!

Prakhar Bindal - 5 years ago

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Could you post your solution too? =D

展豪 張 - 5 years ago

i use the rationalize method to do this

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Could you post your solution too? ;)

展豪 張 - 5 years ago

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don't know how to post here :(

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@Wai Lam William Cheung Credit to @Wai Lam William Cheung :
lim x ( 8 n + 4 n + 2 n 3 2 n ) = lim x 8 n + 4 n + 2 n 3 2 n 1 ( 8 n + 4 n + 2 n ) 2 / 3 + ( 8 n + 4 n + 2 n ) 1 / 3 ( 2 n ) + ( 2 n ) 2 ( 8 n + 4 n + 2 n ) 2 / 3 + ( 8 n + 4 n + 2 n ) 1 / 3 ( 2 n ) + ( 2 n ) 2 = lim x ( 8 n + 4 n + 2 n ) 8 n ( 8 n + 4 n + 2 n ) 2 / 3 + ( 8 n + 4 n + 2 n ) 1 / 3 ( 2 n ) + ( 2 n ) 2 = lim x 4 n + 2 n ( 8 n + 4 n + 2 n ) 2 / 3 + ( 8 n + 4 n + 2 n ) 1 / 3 ( 2 n ) + ( 2 n ) 2 = lim x 1 + ( 1 2 ) n ( 1 + ( 1 2 ) n + ( 1 4 ) n ) 2 / 3 + ( 1 + ( 1 2 ) n + ( 1 4 ) n ) 1 / 3 + 1 = 1 1 + 1 + 1 = 1 3 \;\;\;\;\displaystyle\lim_{x\to\infty}(\sqrt[3]{8^n+4^n+2^n}-2^n) \\ \displaystyle=\lim_{x\to\infty}\dfrac{\sqrt[3]{8^n+4^n+2^n}-2^n}1\dfrac{(8^n+4^n+2^n)^{2/3}+(8^n+4^n+2^n)^{1/3}(2^n)+(2^n)^2}{(8^n+4^n+2^n)^{2/3}+(8^n+4^n+2^n)^{1/3}(2^n)+(2^n)^2} \\ \displaystyle=\lim_{x\to\infty}\dfrac{(8^n+4^n+2^n)-8^n}{(8^n+4^n+2^n)^{2/3}+(8^n+4^n+2^n)^{1/3}(2^n)+(2^n)^2} \\ \displaystyle=\lim_{x\to\infty}\dfrac{4^n+2^n}{(8^n+4^n+2^n)^{2/3}+(8^n+4^n+2^n)^{1/3}(2^n)+(2^n)^2} \\ \displaystyle=\lim_{x\to\infty}\dfrac{1+(\frac 12)^n}{(1+(\frac 12)^n+(\frac 14)^n)^{2/3}+(1+(\frac 12)^n+(\frac 14)^n)^{1/3}+1} \\ =\dfrac 1{1+1+1} \\ =\dfrac 13

展豪 張 - 5 years ago
Chan Lye Lee
Jan 10, 2019

Let x = ( 8 n + 4 n + 2 n ) 1 3 x=\left(8^n+4^n+2^n\right)^{\frac{1}{3}} and y = 2 n = ( 8 n ) 1 3 y=2^n=\left(8^n\right)^{\frac{1}{3}} . Since x 3 y 3 = ( x + y ) ( x 2 + x y + y 2 ) x^3-y^3=(x+y)\left(x^2+xy+y^2\right) , we have x y = x 3 y 3 x 2 + x y + y 2 x-y=\frac{x^3-y^3}{x^2+xy+y^2} , which means that ( 8 n + 4 n + 2 n ) 1 3 2 n = 4 n + 2 n ( 8 n + 4 n + 2 n ) 2 3 + ( 8 n + 4 n + 2 n ) 1 3 × 2 n + 4 n = 1 + ( 1 2 ) n ( 1 + ( 1 2 ) n + ( 1 4 ) n ) 2 3 + 1 + 1 ( 1 + ( 1 2 ) n + ( 1 4 ) n ) 1 3 \left(8^n+4^n+2^n\right)^{\frac{1}{3}} - 2^n = \frac{4^n+2^n}{ \left(8^n+4^n+2^n\right)^{\frac{2}{3}} +\left(8^n+4^n+2^n\right)^{\frac{1}{3}}\times 2^n +4^n } = \frac{1+\left(\frac{1}{2}\right)^n}{ \left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n \right)^{\frac{2}{3}} +1 + 1\left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n\right)^{\frac{1}{3}}}

This means that lim n ( ( 8 n + 4 n + 2 n ) 1 3 2 n ) = lim n 1 + ( 1 2 ) n ( 1 + ( 1 2 ) n + ( 1 4 ) n ) 2 3 + 1 + 1 ( 1 + ( 1 2 ) n + ( 1 4 ) n ) 1 3 = 1 3 \lim_{n \to \infty}\left(\left(8^n+4^n+2^n\right)^{\frac{1}{3}} - 2^n\right) = \lim_{n \to \infty} \frac{1+\left(\frac{1}{2}\right)^n}{ \left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n \right)^{\frac{2}{3}} +1 + 1\left(1+\left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n\right)^{\frac{1}{3}}} =\frac{1}{3}

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