Limits

• August 8th 2010, 05:30 AM
cris
Limits
I need some help to find this limit:

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

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

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

Help?
• August 8th 2010, 05:37 AM
Also sprach Zarathustra
Hint:

3^n < {2^n+3^n} < 2*3^n
• August 8th 2010, 05:38 AM
mr fantastic
Quote:

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

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

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

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

Help?

Note that $\displaystyle \lim_{n\to \infty} \left(2^n+3^n\right)^{\frac{1}{n}} = 3 \lim_{n\to \infty} \left(1 + \left(\frac{2}{3}\right)^n\right)^{\frac{1}{n}}$.
• August 8th 2010, 05:39 AM
Plato
Quote:

Originally Posted by cris
$\lim\limits_{n\to \infty}\((2^n+3^n)^{\frac{1}{n}}$

Suppose that $0 < a < b$ then $b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{a^n + b^n }} \leqslant \sqrt[n]{{b^n + b^n }} \leqslant b\sqrt[n]{2}$

From the above, your limit should be obvious.
• August 8th 2010, 06:45 AM
cris
Quote:

Originally Posted by mr fantastic
Note that $\displaystyle \lim_{n\to \infty} \left(2^n+3^n\right)^{\frac{1}{n}} = 3 \lim_{n\to \infty} \left(1 + \left(\frac{2}{3}\right)^n\right)^{\frac{1}{n}}$.

Is it correct if I say that $(\frac {2}{3})^n$ tends to zero, then this limit is 3?

Quote:

Originally Posted by Plato
Suppose that $0 < a < b$ then $b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{a^n + b^n }} \leqslant \sqrt[n]{{b^n + b^n }} \leqslant b\sqrt[n]{2}$

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?
• August 8th 2010, 06:50 AM
Plato
$0 < a\, \Rightarrow \,\left( {\sqrt[n]{a}} \right) \to 1$
• August 8th 2010, 08:58 AM
HallsofIvy
Quote:

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

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

I tried to start doing something like
$[\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.

Quote:

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

Help?
• August 8th 2010, 03:50 PM
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
• August 8th 2010, 05:28 PM
Plato
Quote:

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?
• August 8th 2010, 06:39 PM
AllanCuz
Quote:

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

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

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

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

Help?

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

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

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

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

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

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

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

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

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

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.

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

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

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 $e^{~0} = 1$ resulting in the answer 3.
• August 8th 2010, 07:12 PM
Also sprach Zarathustra
Quote:

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:

Quote:

One man cannot practice many arts with success.
?