Limit of difference of roots

Calculus Level 5

lim n n ( n 7 + 7 n 6 7 n 2 + 2 n ) = ? \lim_{n\to\infty} n\left(\sqrt[7]{n^7+7n^6} - \sqrt{n^2+2n} \right) = \text{ } \large{?}


The answer is -2.5.

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1 solution

Guilherme Niedu
Jul 13, 2017

L = lim n n [ n 7 + 7 n 6 7 n 2 + 2 n ] \large \displaystyle L = \lim_{n \rightarrow \infty} n \left [\sqrt[7]{n^7+7n^6} - \sqrt{n^2+2n} \right ]

L = lim n n 2 [ ( 1 + 7 n ) 1 7 ( 1 + 2 n ) 1 2 ] \large \displaystyle L = \lim_{n \rightarrow \infty} n^2 \left [ \left(1+\frac7n \right)^{\frac17} - \left(1+\frac2n \right)^{\frac12} \right ]

Using the formula for binomial expansion of ( 1 + x ) p (1+x)^p :

L = lim n n 2 [ ( 1 + 1 7 7 n 3 49 ( 7 n ) 2 + O ( n 3 ) ) ( 1 + 1 2 2 n 1 8 ( 2 n ) 2 + O ( n 3 ) ) ] \large \displaystyle L = \lim_{n \rightarrow \infty} n^2 \left [ \left(1+\frac17 \cdot \frac7n - \frac{3}{49}\cdot \left(\frac 7n \right )^2 + \mathbb{O}(n^{-3}) \right ) - \left(1+ \frac12 \cdot \frac 2n - \frac18 \cdot \left ( \frac 2n \right) ^2 + \mathbb{O}(n^{-3}) \right) \right ]

L = lim n n 2 [ 1 2 n 2 3 n 2 + O ( n 3 ) ] \large \displaystyle L = \lim_{n \rightarrow \infty} n^2 \left [ \frac{1}{2n^2} - \frac{3}{n^2} + \mathbb{O}(n^{-3}) \right]

L = lim n [ 1 2 3 + O ( n 1 ) ] \large \displaystyle L = \lim_{n \rightarrow \infty} \left [ \frac12- 3 + \mathbb{O}(n^{-1}) \right]

L = 5 2 = 2.5 \color{#3D99F6} \boxed{\large \displaystyle L = - \frac52 = -2.5}

Very good! Except it should be ( 7 / n ) 2 (7/n)^2 instead of ( 1 / 7 ) 2 (1/7)^2 .

Ariel Gershon - 3 years, 11 months ago

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Thanks! Fixed it.

Guilherme Niedu - 3 years, 11 months ago

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