Interesting Limit (5)

Calculus Level 3

lim n π n + e n n = ? \large\lim_{n\to\infty}\sqrt[n]{\pi^n+e^n}=\, ?

e e The limit does not exist. π \pi π + e \pi+e

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2 solutions

Brian Lie
Mar 21, 2018

Relevant wiki: Squeeze Theorem

Note that π = π n n π n + e n n π n + π n n = π 2 n \pi=\sqrt[n]{\pi^n}\le \sqrt[n]{\pi^n+e^n}\le \sqrt[n]{\pi^n+\pi^n}=\pi\sqrt[n]2

And lim n π 2 n = π \lim_{n\to\infty}\pi\sqrt[n]2=\pi

By the squeeze theorem, we get lim n π n + e n n = π \lim_{n\to\infty}\sqrt[n]{\pi^n+e^n}=\boxed\pi

This is actually quite intuitive: π n \pi^n grows faster than e n e^n , since π > e \pi>e . Therefore, we can essentially exclude e n e^n from the limit: π n n = π \sqrt[n]{\pi^n}=\boxed{\pi} .

Blan Morrison - 3 years, 2 months ago
Chew-Seong Cheong
Mar 21, 2018

Relevant wiki: L'Hopital's Rule - Basic

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

Haha same.

Arkajyoti Banerjee - 3 years, 2 months ago

I think if you use L 'Hopital's rule, you have to use the Heine theorem. Does n mean positive integers here?Is it real?

(By the way, My English level is limited)

aLIEz βios - 3 years, 2 months ago

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We can use just L'Hopital's rule here. I don't know what Heine theorem is . n n is real and may not necessary be an integer.

Chew-Seong Cheong - 3 years, 2 months ago

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