http://www.mathhelpforum.com/math-he....jpg%5B/IMG%5Dhttp://i131.photobucket.com/albums/p...s/new1/eq2.jpg

how can I find it?

thanks in advance,

T

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- Nov 12th 2007, 05:10 PMTuugiiLooks like a simple limit, help. :)
http://www.mathhelpforum.com/math-he....jpg%5B/IMG%5Dhttp://i131.photobucket.com/albums/p...s/new1/eq2.jpg

how can I find it?

thanks in advance,

T - Nov 12th 2007, 05:34 PMgalactus
Another 'e' limit

- Nov 12th 2007, 06:49 PMTuugii
- Nov 12th 2007, 07:26 PMangel.white
$\displaystyle \lim_{n\to \infty}(n^{\frac{3}{x^{8}}}+\frac{1}{n})^{n}$

Note that this looks strikingly similar to the definition of e:

$\displaystyle \lim_{n \to \infty}(1+\frac{1}{n})^{n}$

So lets see what $\displaystyle n^{\frac{3}{n^{8}}}$ is doing as x goes to infinity. Lets set it equal to some value, we'll choose y to be that value

$\displaystyle y=\lim_{x\to \infty} n^{\frac{3}{n^{8}}}$

Take the natural log in order to get rid of that ugly exponent.

$\displaystyle ln(y)=\lim_{x\to \infty} ln(n^{\frac{3}{n^{8}}})$

By exponential logarithmic identities

$\displaystyle ln(y)=\lim_{x\to \infty} \frac{3}{n^{8}}ln(n)$

You can see that the numerator is going to infinity and the denominator is going to infinity, so we can use L'Hospital's rule

$\displaystyle ln(y)=\lim_{x\to \infty} \frac{3\frac{1}{n}}{8n^{7}}$

Simplify

$\displaystyle ln(y)=\lim_{x\to \infty} \frac{3}{8n^{8}}$

Send x to infinity

$\displaystyle ln(y)=\frac{3}{8\infty^{8}}$

Simplify

$\displaystyle ln(y)=\frac{3}{\infty}$

Simplify

$\displaystyle ln(y)=0$

Solve for y

$\displaystyle y=e^0$

Simplify

$\displaystyle y=1$

Therefore:

$\displaystyle y=\lim_{x\to \infty} n^{\frac{3}{n^{8}}}=1$

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So we know that as x goes to infinity, the first term goes to 1, so let's substitute that into our equation.

$\displaystyle \lim_{n\to \infty}(1+\frac{1}{n})^{n}$

And now we see that as x goes to infinity, the equation goes to the definition of e, so the answer is e. - Nov 12th 2007, 08:12 PMTuugii
awesome!

- Nov 12th 2007, 09:34 PMTuugii
hey wait a minute,

you had found the limit to be 1, but it does not mean the expression without the limit will also equal 1, right? Can you explain this?

thanks,

T - Nov 12th 2007, 09:38 PMangel.white
- Nov 12th 2007, 09:46 PMTuugii
How can you take limits directly, for example:

lim (1+1/n+1/n)^n = e^2 and lim (1+1/logn+1/n)^n = infinity.

I mean, you had clearly showed for the expression that the limit as n approaches 0 will equal to 1, but I was just wondering how it can directly equal to the expression. I understand that the limit is 1 but I don't understand why the expression without the limit is also 1?....

thanks,

T - Nov 12th 2007, 10:48 PMangel.white
Hmm, I can't say for certain, but if I had to guess, I'd say it's because $\displaystyle n^{\frac{3}{n^{8}}}$ goes to 1 quicker than the outside exponent goes to infinity. As n grows, n^8 grows infinitely faster than n, so the outside exponent doesn't affect it.

My book, in the L'Hospital's rule section says "There is a struggle between numerator and denominator. If the numerator wins, the limit will be infinity, if the denominator wins the answer will be zero. Or there may be some compromise, in which case the answer may be some finite positive number."

I think in this case, we have a "struggle" between the function going to 1 and the exponent going to infinity. As you can see, the function going to 1 goes to 1 exponentially quicker than the function going to infinity, so I expect that it "wins" and becomes equal to 1. (if you do this equation with out the +1/n, you get 1 as your answer.)

I expect there is a better way to solve this that doesn't invite such ambiguity, but I can't think of what it would be. - Nov 13th 2007, 06:58 AMThePerfectHacker
$\displaystyle \lim n^{1/n^8} = 1$ because, $\displaystyle 1\leq n^{1/n^{8}} \leq n^{1/n}$ and $\displaystyle \lim n^{1/n} = 1$.

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Best way to solve this problem is to use the fact that $\displaystyle (1+x_n/n)^{n} \to e^x$ where $\displaystyle \lim x_n = x$. - Nov 13th 2007, 08:10 AMTuugii
- Nov 13th 2007, 11:00 AMangel.white
I think I can, so I'll give it a shot (feel free to correct me PH, if I err)

$\displaystyle \lim_{n\to\infty}\left( 1\leq n^{3/n^{8}} \leq n^{1/n}\right)$

First lets show that

$\displaystyle \lim_{n\to\infty}\left(1\leq n^{3/n^{8}}\right)$

Because n is going to infinity, the base, n, will always be greater than 1. And the only way to make a number greater than 1 be less than 1 with exponents is to take it to a negative power. However $\displaystyle \lim_{n\to\infty}\frac{3}{n^{8}}$ can never be negative, because 3 is positive, and n is positive (and a number taken to an even power is always positive). So you have a base always greater than one equal to an exponent which is always positive, so this value will never be less than one. We know that it can be equal to one because any positive number taken to the zero power is 1, so if the exponent equals zero, then the limit equals one.

Now for

$\displaystyle \lim_{n\to\infty}\left( n^{3/n^{8}} \leq n^{1/n}\right)$

$\displaystyle \lim_{n\to\infty}\left( \frac{3}{n^{8}}ln(n)\leq \frac{1}{n}ln(n)\right)$

$\displaystyle \lim_{n\to\infty}\left( \frac{3}{n^{8}}\leq \frac{1}{n}\right)$

$\displaystyle \lim_{n\to\infty}\left( \frac{3}{n^{7}}\leq 1\right)$

Which, we can see is true, so because the base is the same and the exponent goes to zero quicker on the LHS, the LHS can never be greater than the RHS.

And now $\displaystyle \lim_{n\to\infty} n^{\frac{1}{n}} = 1$

$\displaystyle \lim_{n\to\infty} n^{\frac{1}{n}} = y$

$\displaystyle \lim_{n\to\infty} \frac{1}{n}\ln(n) = ln(y)$

L'Hospital's Rule

$\displaystyle \lim_{n\to\infty} \frac{\frac{1}{n}}{1} = ln(y)$

$\displaystyle \lim_{n\to\infty} \frac{1}{n} = ln(y)$

$\displaystyle \frac{1}{\infty} = ln(y)$

$\displaystyle 0 = ln(y)$

$\displaystyle e^{0} = y$

$\displaystyle 1 = y = \lim_{n\to\infty} n^{\frac{1}{n}}$

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**CONCLUSION:**

So we have shown that $\displaystyle \lim_{n\to\infty}\left( 1\leq n^{3/n^{8}}\right)$

And that $\displaystyle \lim_{n\to\infty}\left(n^{3/n^{8}} \leq n^{1/n}\right)$

And thus $\displaystyle \lim_{n\to\infty}\left( 1\leq n^{3/n^{8}} \leq n^{1/n}\right)$

And we have shown $\displaystyle \lim_{n\to\infty} n^{\frac{1}{n}} = 1$

Thus

And thus $\displaystyle \lim_{n\to\infty}\left( 1\leq n^{3/n^{8}} \leq 1\right)$

So $\displaystyle \lim_{n\to\infty}n^{3/n^{8}}$ must equal one.

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**PROBLEMS:**Now that I am done, I think I did a good job showing that the limit must be n, but I am confused by how this means it is not affected by the outside going to infinity. For example, I could use the same method to prove $\displaystyle \lim_{n\to\infty} \frac{1}{n}=0$ but we can plainly see in this equation that it is affected by being taken to the power of infinity.

As an example, I entered $\displaystyle \lim_{n\to \infty}\left(n^{\frac{3}{x^{8}}}+\frac{1}{n}\right )^{n}$ into Function calculator and it agreed that the limit is e, but when I changed the equation to $\displaystyle \lim_{n\to \infty}\left(n^{\frac{3}{x}}+\frac{1}{n}\right)^{n }$ the limit became infinity, even though we know that $\displaystyle \lim_{n\to\infty} n^{3/n}=1$

This would seem to seem to indicate that it is subject to being effected by being taken to the power of infinity, but that it simply overpowers this effect due to it's large exponent (being taken to the power of 8), which causes it to go to 1 faster than the exponent can affect it.

So I guess my problem is that I don't understand how my proof shows that it's limit can be found independently of the rest of the equation, when it is nested within another function which could effect it's result. - Nov 13th 2007, 12:02 PMTuugii
I can see it is fairly easy to show the limit to be 1 for (n^3)^(1/n^3) as n approaches infinity. But the problem is that, how can you say the limit of the whole expression is e, as you only know that the limit of the (n^3)^(1/n^3) is 1.

I am wondering how you can take limits separately.?...

thanks,

T - Nov 13th 2007, 02:05 PMThePerfectHacker
I used the

**squeeze theorem**.