# Did i solved it correctly

1. Dec 4, 2008

### transgalactic

lim (2^n + 3^n)^0.25 =3^n[(2/3)^n + 1]^n =+infinity
n->+infinity

2. Dec 4, 2008

### Office_Shredder

Staff Emeritus
Your equation is incorrect.... the first part has a ^(.25) in it which seems to have been replaced by a ^n in the second one. Either way your answer is correct incidentally, but simply factoring out a 3n isn't really enough to be a convincing argument

3. Dec 4, 2008

### Дьявол

You should solve it like this:
$$\lim_{n \rightarrow \infty}(2^n + 3^n)^{0.25} =\lim_{n \rightarrow \infty}\sqrt[4]{2^n + 3^n}$$
Now, you need to multiply by:
$$\frac{\sqrt[4]{(2^n+3^n)^3}}{\sqrt[4]{(2^n+3^n)^3}}$$
and you will get what?

4. Dec 4, 2008

### transgalactic

if i will multiply i will get
(2^n + 3^n)/(2^n + 3^n)^0.25

the nuenator is bigger it goes to infinity

am i correct
??

5. Dec 4, 2008

### Tedjn

No, don't forget that the denominator is no longer raised to the 0.25 power. Once you get the right fraction, you could probably expand the denominator and try to simplify the resulting mess. Maybe there is a cleaner way from there, I don't know.

One thing you can do to make this limit essentially clear is to find a similar but always smaller sequence that also tends to infinity as n becomes large. Seek a way to simplify what is inside the parentheses: 2n + 3n. Is there a way to get a sequence in which every term is smaller than the current sequence, which also allows you to easily combine the exponentials?

6. Dec 5, 2008

### Дьявол

I think this is not so hard to solve.
$$\lim_{n \rightarrow \infty}(2^n + 3^n)^{0.25} =\lim_{n \rightarrow \infty}\sqrt[4]{2^n + 3^n} * \frac{\sqrt[4]{(2^n+3^n)^3}}{\sqrt[4]{(2^n+3^n)^3}} =\lim_{n \rightarrow \infty}\frac{2^n+3^n}{\sqrt[4]{(2^n+3^n)^3}}$$
Now expand the denominator using (A+B)3=A3+3A2B+3AB2+B2

7. Dec 5, 2008

### transgalactic

i dont need to expand
if i take (2^n + 3^n)=t

the numerator has power 1
the denominator has power 3/4

the numerator grows faster
thats why the limit goes to infinity

8. Dec 5, 2008

### Дьявол

If you want to prove that the limit goes to infinity you need to write like this:
$$\lim_{t \rightarrow \infty}\frac{t}{\sqrt[4]{t^3}}=\lim_{t \rightarrow \infty}\frac{1}{\sqrt[4]{\frac{1}{t}}}=\frac{1}{0}=\infty$$

Regards.

9. Dec 5, 2008

### JG89

$$2^n < 2^n + 3^n$$ for all n. So $$\sqrt[4](2^n) < \sqrt[4](2^n + 3^n)$$.

Now, $$\sqrt[4](2^n) = 2^(\frac{n}{4})$$ (That says 2^(n/4)). The $$\frac{n}{4}$$ goes to infinity, and so the entire term goes to infinity. Since it is less than the original sequence, what can you say about the limit of the original sequence?