Limit by Squeeze Theorem - Part 1

Calculus Level 2

Find lim n ( n 2 + 1 8 n + 1 4 ) \displaystyle \lim_{n \to \infty} \left(\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right) by squeeze theorem .


The answer is 0.

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1 solution

Chris Lewis
Oct 20, 2020

( n 2 + 0 ) 1 8 ( n + 1 ) 1 4 < ( n 2 + 1 ) 1 8 ( n + 1 ) 1 4 < ( n 2 + 2 n + 1 ) 1 8 ( n + 1 ) 1 4 n 1 4 ( n + 1 ) 1 4 < ( n 2 + 1 ) 1 8 ( n + 1 ) 1 4 < ( n + 1 ) 1 4 ( n + 1 ) 1 4 \begin{aligned} \left( n^2+{\color{#3D99F6} 0} \right)^\frac18 - (n+1)^\frac14 &< \left( n^2 +1 \right)^\frac18 - (n+1)^\frac14 < \left( n^2 {\color{#3D99F6} +2n}+1 \right)^\frac18 - (n+1)^\frac14 \\ n^\frac14 - (n+1)^\frac14 &< \left( n^2 +1 \right)^\frac18 - (n+1)^\frac14 < (n+1)^\frac14 - (n+1)^\frac14 \end{aligned}

Since the right-hand side is zero for all n n , and the left-hand side tends to zero as n n \to \infty , by the squeeze theorem

lim n ( n 2 + 1 ) 1 8 ( n + 1 ) 1 4 = 0 \lim_{n \to \infty} \left( n^2 +1 \right)^\frac18 - (n+1)^\frac14 = \boxed0

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