# Compute the limit (with variable in exponent)

• PirateFan308
So I get \frac{(-2)^{n}(1/3)^{n+1}+(1/3)}{(-2/3)^{n+1}+1}Since (-2)n → +∞; (1/3)n+1 → 0, then the product of the two will converge to 0 and the top will converge to \frac{1}{3}(-2/3)n+1 → 0 so the bottom will converge to 1Yes.

## Homework Statement

Compute the following sequence (show your work)

(-2)n+3n
÷ (-2)n+1+3n+1

## The Attempt at a Solution

(-2)n+3n
÷ (-2)n(-2)+3n3n

After this I don't know what to do. I know that if I can prove that something smaller than this for all n's converges to +∞, this will converge to +∞. Also, if I can prove that something bigger that this for all n's converges to 0, this will converge to 0. I'm just not sure how to simplify this equation at all.

Is the problem:

$$lim_{n \to \infty} \frac{(-2)^{n}+3^{n}}{(-2)^{n+1}+3^{n+1}}$$

?

PirateFan308 said:
(-2)n+3n
÷ (-2)n(-2)+3n3n

Look at this again.

PirateFan308 said:

## Homework Statement

Compute the following sequence (show your work)

(-2)n+3n
÷ (-2)n+1+3n+1

## The Attempt at a Solution

(-2)n+3n
÷ (-2)n(-2)+3n3n

After this I don't know what to do. I know that if I can prove that something smaller than this for all n's converges to +∞, this will converge to +∞. Also, if I can prove that something bigger that this for all n's converges to 0, this will converge to 0. I'm just not sure how to simplify this equation at all.

I'm guess that this is the expression (it's not an equation) whose limit you want to find:
$$\frac{(-2)^n + 3^n}{(-2)^{n+1} + 3^{n+1}}$$

Multiply the numerator by 1/3n+1 and multiply the denominator by the same thing.

Mark44 said:
I'm guess that this is the expression (it's not an equation) whose limit you want to find:
$$\frac{(-2)^n + 3^n}{(-2)^{n+1} + 3^{n+1}}$$

Multiply the numerator by 1/3n+1 and multiply the denominator by the same thing.

So I get $\frac{(-2)^{n}(1/3)^{n+1}+(1/3)}{(-2/3)^{n+1}+1}$

Since (-2)n → +∞; (1/3)n+1 → 0, then the product of the two will converge to 0 and the top will converge to $\frac{1}{3}$

(-2/3)n+1 → 0 so the bottom will converge to 1

So the whole thing will converge to $\frac{1}{3}$

Is this correct?

PirateFan308 said:
So I get $\frac{(-2)^{n}(1/3)^{n+1}+(1/3)}{(-2/3)^{n+1}+1}$

Since (-2)n → +∞; (1/3)n+1 → 0, then the product of the two will converge to 0 and the top will converge to $\frac{1}{3}$
Your reasoning is flawed here. If one factor in a product is growing large without bound (i.e., approaching ∞) and the other factor is approaching 0, there is nothing you can say about the product. The indeterminate form [0* ∞] is similar to other indeterminate forms such as [0/0] and [∞/∞].

For example, would you say that 4n * (1/3)n also has a limit of zero?

Instead of writing (-2)n * (1/3)n+1, write these together as a single fraction, and then you can take the limit.
PirateFan308 said:
(-2/3)n+1 → 0 so the bottom will converge to 1
Yes.
PirateFan308 said:
So the whole thing will converge to $\frac{1}{3}$

Is this correct?