a nice limit : part 2

Calculus Level 3

lim n ( 1 + n 2 ) 1 ln ( n ) = ? \Large \lim_{n \to \infty} (1+n^2)^{\frac 1{\ln(n)}} = \ ?

0 e^2 +\infty ln(2)

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2 solutions

Relevant wiki: L'Hopital's Rule - Basic

L = lim n ( 1 + n 2 ) 1 ln ( n ) = lim n exp ( ln ( ( 1 + n 2 ) 1 ln ( n ) ) ) where exp ( x ) = e x = exp ( lim n ln ( 1 + n 2 ) ln ( n ) ) A / case, L’H o ˆ pital’s rule applies. = exp ( lim n 2 n 1 + n 2 1 n ) Differentiate up and down w.r.t. n . = exp ( lim n 2 n 2 1 + n 2 ) Divide up and down by n 2 . = exp ( lim n 2 1 n 2 + 1 ) = e 2 \begin{aligned} L & = \lim_{n \to \infty}(1+n^2)^{\frac 1{\ln(n)}} \\ & = \lim_{n \to \infty} \exp \left(\ln \left((1+n^2)^{\frac 1{\ln(n)}}\right) \right) & \small \color{#3D99F6} \text{where }\exp (x) = e^x \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln (1+n^2)}{\ln(n)} \right) & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \exp \left(\lim_{n \to \infty} \frac {\frac {2n}{1+n^2}}{\frac 1n} \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }n. \\ & = \exp \left(\lim_{n \to \infty} \frac {2n^2}{1+n^2} \right) & \small \color{#3D99F6} \text{Divide up and down by }n^2. \\ & = \exp \left(\lim_{n \to \infty} \frac 2{\frac 1{n^2}+1} \right) \\ & = \boxed{e^2} \end{aligned}

Ahpa Tsum
Aug 7, 2018

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