I'm interested in the proofs of the following: Using the definition of limit (for sequences).
I proved the first one, as shown, but I don't know how to go about doing the second and third.
- $\displaystyle \lim_{n \to \infty} \frac{n}{3^n} = 0$. Proof: Let $\epsilon > 0$. If $n > M$, then $\vert \frac{n}{3^n} - 0\vert = \frac{\displaystyle n}{\displaystyle \Big(\frac32\Big)^n2^n} \leq \frac{\displaystyle n}{\displaystyle (1 + \frac{n}2)(1 + n)} \leq \frac{\displaystyle n}{\displaystyle (1 + n/2)n} = \frac{1}{\displaystyle 1 + \frac12 n} \leq \frac{1}{\frac12 n} = \frac2n < \epsilon$ when $\displaystyle M = \frac{2}{\epsilon}$.
But for these I can't seem to get them. Please provide a proof. Or a good hint! Thanks!
$\displaystyle \lim_{n \to \infty} \frac{n^3}{2^n} = 0$.
$\displaystyle \lim_{n \to \infty} \frac{n^6}{3^n} = 0$.