Hey, I am trying to prove that:

lim(nth root of n)=1 as n approaches infinity and that

lim(nth root of n+1)=1 as n approaces infinity

I do not know how to go about it. Can anybody help??

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- Jan 6th 2006, 03:06 PMTexasGirlLimit Proof
Hey, I am trying to prove that:

lim(nth root of n)=1 as n approaches infinity and that

lim(nth root of n+1)=1 as n approaces infinity

I do not know how to go about it. Can anybody help?? - Jan 6th 2006, 04:31 PMJameson
Is using the delta-epsilon definition of a limit or a less rigorous method?

Start with considering any function $\displaystyle f(n)=n^\left(\frac{1}{n}\right)$. You can do this both graphically and analytically. For the first way, use a calculator to see the trend as n tends to infinity. For the second way, look at the exponential part of the function. What is $\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n}$? And what is any number or variable to that answer? - Jan 6th 2006, 06:18 PMThePerfectHacker
This happens to be a famous limit

I will prove that $\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}$=1.

Notice that the function $\displaystyle n^{\frac{1}{n}}$ is a monotonic decreasing function, which has a lower bound. Thus, $\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}$ has a limit. Call that $\displaystyle L$ thus,

L=$\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}$

thus,

$\displaystyle \ln L=\ln (\lim_{n\rightarrow \infty} n^{\frac{1}{n}})$

But because $\displaystyle \ln$ is countinous for that interval,

$\displaystyle \ln L=\lim_{n\rightarrow \infty} \ln (n^{\frac{1}{n}})$

Thus,

$\displaystyle \ln L=\lim_{n\rightarrow \infty} \frac{\ln n}{n}$

But this fits the necessary condition of L'Hopital's Rule,

$\displaystyle \ln L=\lim_{n\rightarrow \infty} \frac{1}{n}=0$

Thus, $\displaystyle \ln L=0$ thus, $\displaystyle L=1$

Finally putting all of this together,

$\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}$=1 - Jan 6th 2006, 06:21 PMThePerfectHackerQuote:

Originally Posted by**Jameson**

- Jan 6th 2006, 11:25 PMCaptainBlackQuote:

Originally Posted by**Jameson**

being raised to $\displaystyle 1/n$. Such an argument would apply to

$\displaystyle \lim_{n\rightarrow\infty}f(n)^{\frac{1}{n}}$

for any real function $\displaystyle f(n)$.

In particular it would appy when $\displaystyle f(n)=n^n$.

RonL - Jan 6th 2006, 11:27 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

RonL - Jan 7th 2006, 01:28 AMCaptainBlackQuote:

Originally Posted by**TexasGirl**

$\displaystyle n^{\frac{1}{n}}=e^{\frac{ \ln(n)}{n}}$,

and as the exponential function is continuous if the limit exists:

$\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=e^{\lim_{n\rightarrow \infty}\frac{ \ln(n)}{n}}$,

so we need only worry about:

$\displaystyle \lim_{n\rightarrow \infty}\frac{ \ln(n)}{n}$.

Now this is dealt with elsewhere in this thread by ThePerfectHacker, but

an informal short cut is available which tells us what this limit is.

The sort cut is that log functions grow more slowly than any power

of its argument, that is for all $\displaystyle \alpha \epsilon \mathbb{R}$, there exists a $\displaystyle k$ such that:

$\displaystyle \ln (x) < k.x^{\alpha}$,

for all $\displaystyle x$ sufficiently large. Which is sufficient to show that

$\displaystyle \lim_{n\rightarrow \infty}\frac{ \ln(n)}{n^{\beta}}=0$,

for any $\displaystyle \beta \epsilon \mathbb{R}$. Of course this can also be demonstrated

using L'Hopital's rule as is done for $\displaystyle \beta = 1$ elsewhere in this thread.

RonL - Jan 7th 2006, 07:59 AMTexasGirlMany Thanks
You guys are awesome. I don't know what I would do without you. Thanks a million!

- Jan 7th 2006, 11:27 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

- Jan 7th 2006, 12:24 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

growth of what is raised to $\displaystyle 1/n$ not just that the limit

of the exponent is $\displaystyle 0$. Its explained in my response to

Jameson's post.

RonL - Jan 7th 2006, 02:11 PMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

For the second thing you cannot just take the natural logarithm of the limit you must first prove it exists. It is a common mistake by just taking the natural logarithm because we are assuming it exits. - Jan 8th 2006, 08:18 AMJamesonQuote:

Originally Posted by**CaptainBlack**

- Jan 8th 2006, 08:24 AMJamesonQuote:

Originally Posted by**ThePerfectHacker**

- Jan 8th 2006, 09:06 AMCaptainBlackQuote:

Originally Posted by**Jameson**

**if**valid would also prove that:

$\displaystyle \lim_{x\rightarrow\infty}(n^n)^{1/n}=1$,

as here also the outer exponent goes to zero and as we know anything

raised to the power zero (if non-zero) is unity.

RonL - Jan 8th 2006, 09:38 AMThePerfectHacker
Jameson, your problem, informally was that you looked at the exponent you should have also seen the base thus it is of the form $\displaystyle \infty^0$ no conclusion could be drawn.