Compute the limit $\lim\limits_{n\to\infty}a_n$ for the following sequences:
(a) $a_n=e^{5\cos((\pi/6)^n)}$
(b) $a_n=\frac{n!}{n^n}$
For part (a) do I just take the limit of the exponent part and then the answer would be $e$ raised to whatever the limit is?
And would the limit be $1$ or $-1$? because $\cos$ goes between those two.
For part $b$ it is in the form of infinity over infinity but how do you take the derivative of $n!$? Will it ever break out of infinity over infinity?
Best Answer
(a) $$ \lim_{n \to \infty} e^{5 \cos((\frac{\pi}6))^n} = e^{\lim_{n \to \infty} 5 \cos((\frac{\pi}6))^n} = e^{5 \cos(\lim_{n \to \infty}(\frac{\pi}6)^n)} = e^{5 \cdot 1} = e^5 $$
(b) $$ \frac{n!}{n^n} = \frac{1\cdot 2 \cdots n}{n\cdot n \cdot n} \leq \frac{1}{n} $$ From here $$ \lim_{n \to \infty} \frac{n!}{n^n} = 0 $$ since $$ \lim_{n \to \infty} \frac{1}{n} = 0 $$