Limited

Calculus Level 3

lim ⁡ n → ∞ a 2 n + n ! ( a + 1 ) n + ( n + 1 ) ! = ? \large\displaystyle\lim_{n\rightarrow \infty}\frac{a^{2n}+n!}{(a+1)^n+(n+1)!}=\ ?

Take a a as a positive real constant.


The answer is 0.

Aleksa Radovanović by Aleksa Radovanović , 25, Serbia

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1 solution

Adrian Castro
Jan 9, 2016

Let e = a 2 e=a^2 and f = a + 1 f=a+1

Then it follows: lim ⁡ n → ∞ e n + n ! f n + ( n + 1 ) ! = lim ⁡ n → ∞ e n + n ! f n + ( n + 1 ) ! ⋅ 1 ( n + 1 ) ! 1 ( n + 1 ) ! = lim ⁡ n → ∞ e n ( n + 1 ) ! + 1 n + 1 f n ( n + 1 ) ! + 1 \lim_{n\rightarrow{\infty}}\frac{e^n+n!}{f^n+(n+1)!}=\lim_{n\rightarrow{\infty}}\frac{e^n+n!}{f^n+(n+1)!}\cdot\frac{\frac{1}{(n+1)!}}{\frac{1}{(n+1)!}} =\lim_{n\rightarrow{\infty}}\frac{\frac{e^n}{(n+1)!}+\frac{1}{n+1}}{\frac{f^n}{(n+1)!}+1}

Since lim ⁡ n → ∞ x n n ! = 0 ∣ x ϵ R \lim_{n\rightarrow{\infty}}\frac{x^n}{n!}=0\mid x\epsilon \mathbb{R} our answer is 0 \boxed{0}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

That's it :D

Aleksa Radovanović - 5 years, 5 months ago

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