Compute the limit (with variable in exponent)

[PirateFan308]

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.

 

[gb7nash]

Is the problem:

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

?

(-2)ⁿ+3ⁿ
÷ (-2)ⁿ(-2)+3ⁿ3ⁿ

Look at this again.

 

[Mark44]

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.

 

[PirateFan308]

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?

 

[Mark44]

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.

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

Yes.

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

Is this correct?