1. ## Limits at infinity

Hey, this question is simple enough, I know the answer. But I need a justification for it.

As x -> infinity, f(x) = xa^x -> 0. ( if 0 < a < 1)

I can justify that a^x goes to 0 when x goes to infinity, but that other x term is making it a little more complicated to justify. Do i just say that a^x is getting smaller at a rate faster than x is increasing?

2. Originally Posted by hashshashin715
Hey, this question is simple enough, I know the answer. But I need a justification for it.

As x -> infinity, f(x) = xa^x -> 0. ( if 0 < a < 1)

I can justify that a^x goes to 0 when x goes to infinity, but that other x term is making it a little more complicated to justify. Do i just say that a^x is getting smaller at a rate faster than x is increasing?
$\displaystyle 0 \times \text{anything} = 0$

so

$\displaystyle 0 \times \infty = 0$

3. Are you allowed to use the ratio test?

$\displaystyle \lim_{n \to \infty} \frac{(n+1)a^{(n+1)}}{na^n} =$

$\displaystyle \lim_{n \to \infty} \frac{(n+1)}{n} \cdot \frac{a^{(n+1)}}{a^n} =$

$\displaystyle \lim_{n \to \infty} \frac{(n+1)}{n} \cdot \lim_{n \to \infty} \frac{a^{(n+1)}}{a^n} =$

$\displaystyle 1 \cdot a = a$

And since $\displaystyle 0 < a < 1$,

$\displaystyle \lim_{n \to \infty} na^n = 0$

4. Originally Posted by pickslides

$\displaystyle 0 \times \infty = 0$
Not true in general.