1. ## Limits

I need some help to find this limit:

$\displaystyle$\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$$I tried to start doing something like \displaystyle [\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}$$

But then I don't know what to do with this $\displaystyle 2^n$ and $\displaystyle 3^n$

Help?

2. Hint:

3^n < {2^n+3^n} < 2*3^n

3. Originally Posted by cris
I need some help to find this limit:

$\displaystyle$\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$$I tried to start doing something like \displaystyle [\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}$$

But then I don't know what to do with this $\displaystyle 2^n$ and $\displaystyle 3^n$

Help?
Note that $\displaystyle \displaystyle \lim_{n\to \infty} \left(2^n+3^n\right)^{\frac{1}{n}} = 3 \lim_{n\to \infty} \left(1 + \left(\frac{2}{3}\right)^n\right)^{\frac{1}{n}}$.

4. Originally Posted by cris
$\displaystyle$\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$$Suppose that \displaystyle 0 < a < b then \displaystyle b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{a^n + b^n }} \leqslant \sqrt[n]{{b^n + b^n }} \leqslant b\sqrt[n]{2} From the above, your limit should be obvious. 5. Originally Posted by mr fantastic Note that \displaystyle \displaystyle \lim_{n\to \infty} \left(2^n+3^n\right)^{\frac{1}{n}} = 3 \lim_{n\to \infty} \left(1 + \left(\frac{2}{3}\right)^n\right)^{\frac{1}{n}}. Is it correct if I say that \displaystyle (\frac {2}{3})^n tends to zero, then this limit is 3? Originally Posted by Plato Suppose that \displaystyle 0 < a < b then \displaystyle b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{a^n + b^n }} \leqslant \sqrt[n]{{b^n + b^n }} \leqslant b\sqrt[n]{2} From the above, your limit should be obvious. I understood what you did... and I would think about doing 0 < a < b, but this second part... Did you just deduce that? Or is it a theorem I should know? 6. \displaystyle 0 < a\, \Rightarrow \,\left( {\sqrt[n]{a}} \right) \to 1 7. Originally Posted by cris I need some help to find this limit: \displaystyle \lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$$

I tried to start doing something like
$\displaystyle$[\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}$$Actually, this makes no sense. You can't has a power of 1/n outside the limit as n goes to infinity. But then I don't know what to do with this \displaystyle 2^n and \displaystyle 3^n Help? 8. 3^n < {2^n+3^n} < 2*3^n ==> {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2*3^n}^{1/n} ==> {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2}^{1/n} * {3^n}^{1/n} ==> 3 < {2^n+3^n}^{1/n} < {2}^{1/n} * 3 ==> lim{3} <lim[ {2^n+3^n}^{1/n}] < lim[{2}^{1/n} * 3 ] ==> 3 <lim[ {2^n+3^n}^{1/n}] < 3 lim[{2}^{1/n} ] { lim 2^1/n =1 } ==> 3 <lim[ {2^n+3^n}^{1/n}] < 3 Applying sandwich rule : lim[ {2^n+3^n}^{1/n}] = 3 9. Originally Posted by Also sprach Zarathustra 3^n < {2^n+3^n} < 2*3^n ==> {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2*3^n}^{1/n} ==> {3^n}^{1/n} < {2^n+3^n}^{1/n} < {2}^{1/n} * {3^n}^{1/n} ==> 3 < {2^n+3^n}^{1/n} < {2}^{1/n} * 3 ==> lim{3} <lim[ {2^n+3^n}^{1/n}] < lim[{2}^{1/n} * 3 ] ==> 3 <lim[ {2^n+3^n}^{1/n}] < 3 lim[{2}^{1/n} ] { lim 2^1/n =1 } ==> 3 <lim[ {2^n+3^n}^{1/n}] < 3 Applying sandwich rule : lim[ {2^n+3^n}^{1/n}] = 3 What a mess that post is! I am willing to bet that were Nietzsche alive today he would have embraced the use of LaTeX. Why don’t you learn to use it? 10. Originally Posted by cris I need some help to find this limit: \displaystyle \lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$$

I tried to start doing something like
$\displaystyle$[\lim\limits_{n\to \infty}\((2^n+3^n)]^{\frac{1}{n}}

But then I don't know what to do with this $\displaystyle 2^n$ and $\displaystyle 3^n$

Help?
Let $\displaystyle y = (2^n + 3^n )^{ \frac{1}{n} }$

$\displaystyle lny = \frac{1}{n} ln (2^n + 3^n)$

$\displaystyle \lim\limits_{ n \to \infty } y = \lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln (2^n + 3^n) }$

$\displaystyle \lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln (3^n [ 1 + ( \frac{2}{3} )^n ] ) }$

$\displaystyle \lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln (3^n ) + \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ] }$

$\displaystyle \lim\limits_{ n \to \infty } e^{ n \frac{1}{n} ln (3 ) + \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ] }$

$\displaystyle 3\lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ] }$

$\displaystyle 3\lim\limits_{ n \to \infty } e^{ ln [ 1 + ( \frac{2}{3} )^n ]^{ \frac{1}{n} }$

$\displaystyle 3\lim\limits_{ n \to \infty } [ 1 + ( \frac{2}{3} )^n ]^{ \frac{1}{n}$

Wow. And after all that I end up at Mr. Fantastics result. I'm actually depressed.

In any event, to evaluate Mr. Fantastics limit we will revert one step.

$\displaystyle 3\lim\limits_{ n \to \infty } e^{ ln [ 1 + ( \frac{2}{3} )^n ]^{ \frac{1}{n} }$

$\displaystyle 3\lim\limits_{ n \to \infty } e^{ \frac{1}{n} ln [ 1 + ( \frac{2}{3} )^n ] }$

Note how the 1 over n will tend towards 0 as n goes to infinity as will what is in the ln portion of the equation. We will end up with $\displaystyle e^{~0} = 1$ resulting in the answer 3.

11. What a mess that post is!
I am willing to bet that were Nietzsche alive today he would have embraced the use of LaTeX.
Why don’t you learn to use it?
Dear Plato,

Do you know who said:

One man cannot practice many arts with success.
?