Limit of 2^(3n)/3^(2n) as n approaches infinity

[quasar987]

$$\lim_{n\rightarrow \infty} \frac{2^{3n}}{3^{2n}}$$

The answer is zero. All I can do is turn this infinity/infinity undeterminate form into a 0 times infinity indeterminate form. I also tried finding a creature strictly bigger than [itex]\frac{2^{3n}}{3^{2n}}[/tex] that has zero for a limit so that the answer would follow from the "sandwich theorem". But all my attempts let to infinity. For instance,<br /> <br /> $$0\leq \frac{2^{3n}}{3^{2n}}\leq \frac{3^{3n}}{3^{2n}}=\frac{3^{2n}3^n}{3^{2n}}=3^n$$[/itex]

 

[Muzza]

2^(3n) = (2^3)^n = 8^n and 3^(2n) = (3^2)^n = 9^n, so 2^(3n) / 3^(2n) = what?

 

[quasar987]

$$\lim_{n\rightarrow \infty} \left( \frac{8}{9} \right)^n=0$$

Thanx a bunch!

 

[Muzza]

Btw, $$\infty$$ is written \infty in LaTeX.

 

[quasar987]

I would have another question for you if you don't mind.

The textbook says that

$$\lim_{n\rightarrow \infty} \sqrt{2n+1} - \sqrt{2n}=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{2n+1}+\sqrt{2n}}$$

What are the steps to get from one to the other?

 

[Muzza]

Multiply $$\sqrt{2n + 1} - \sqrt{2n}$$ by its conjugate like so:

$$\sqrt{2n + 1} - \sqrt{2n} = (\sqrt{2n + 1} - \sqrt{2n}) \cdot \frac{\sqrt{2n + 1} + \sqrt{2n}}{\sqrt{2n + 1} + \sqrt{2n}}$$.

The top can then be simplified into 1...

 

[quasar987]

Great!

Those were #1 a) and b) out of n) by the way. Wish me luck!

 

[courtrigrad]

Is this from Courant's book?

 

[quasar987]

No, it's from two Canadian people: Jaques Labelle and Armel Mercier.

 

[quasar987]

I'm using this somewhat dead thread to ask: What does QED means? D is probably for Demonstrated, but what about Q and E?

In french, it's CQFD, which translate into "That which had to be demonstrated."

 

[Tide]

Close! It means "that which was to be shown" or "demonstrated."