# Limit Proof

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• Jan 6th 2006, 04:06 PM
TexasGirl
Limit 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, 05:31 PM
Jameson
Is using the delta-epsilon definition of a limit or a less rigorous method?

Start with considering any function $f(n)=n^\left(\frac{1}{n}\right)$. 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 $\lim_{n\rightarrow\infty}\frac{1}{n}$? And what is any number or variable to that answer?
• Jan 6th 2006, 07:18 PM
ThePerfectHacker
This happens to be a famous limit
I will prove that $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}$=1.
Notice that the function $n^{\frac{1}{n}}$ is a monotonic decreasing function, which has a lower bound. Thus, $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}$ has a limit. Call that $L$ thus,
L= $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}$
thus,
$\ln L=\ln (\lim_{n\rightarrow \infty} n^{\frac{1}{n}})$
But because $\ln$ is countinous for that interval,
$\ln L=\lim_{n\rightarrow \infty} \ln (n^{\frac{1}{n}})$
Thus,
$\ln L=\lim_{n\rightarrow \infty} \frac{\ln n}{n}$
But this fits the necessary condition of L'Hopital's Rule,
$\ln L=\lim_{n\rightarrow \infty} \frac{1}{n}=0$
Thus, $\ln L=0$ thus, $L=1$
Finally putting all of this together,
$\lim_{n\rightarrow \infty} n^{\frac{1}{n}}$=1
• Jan 6th 2006, 07:21 PM
ThePerfectHacker
Quote:

Originally Posted by Jameson
Is using the delta-epsilon definition of a limit or a less rigorous method?

Start with considering any function $f(n)=n^\left(\frac{1}{n}\right)$. 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 $\lim_{n\rightarrow\infty}\frac{1}{n}$? And what is any number or variable to that answer?

Jameson I believe the rule goes that if the composition of the limits exists then the limit is the functional composition. That means what you said about looking at the exponent part lacks mathematical rigor.
• Jan 7th 2006, 12:25 AM
CaptainBlack
Quote:

Originally Posted by Jameson
For the second way, look at the exponential part of the function. What is $\lim_{n\rightarrow\infty}\frac{1}{n}$? And what is any number or variable to that answer?

This is invalid as it pays no attention to the rate of growth of what is
being raised to $1/n$. Such an argument would apply to

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

for any real function $f(n)$.

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

RonL
• Jan 7th 2006, 12:27 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Jameson I believe the rule goes that if the composition of the limits exists then the limit is the functional composition. That means what you said about looking at the exponent part lacks mathematical rigor.

- its actualy false (fallacious?), see my other post :D

RonL
• Jan 7th 2006, 02:28 AM
CaptainBlack
Quote:

Originally Posted by TexasGirl
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??

Now:

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

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

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

so we need only worry about:

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

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 $\alpha \epsilon \mathbb{R}$, there exists a $k$ such that:

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

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

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

for any $\beta \epsilon \mathbb{R}$. Of course this can also be demonstrated
using L'Hopital's rule as is done for $\beta = 1$ elsewhere in this thread.

RonL
• Jan 7th 2006, 08:59 AM
TexasGirl
Many Thanks
You guys are awesome. I don't know what I would do without you. Thanks a million!
• Jan 7th 2006, 12:27 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
- its actualy false (fallacious?), see my other post :D

RonL

What are you talking about CaptainBlack.
• Jan 7th 2006, 01:24 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
What are you talking about CaptainBlack.

Jameson's statement is a fallacy - you need to consider the rate of
growth of what is raised to $1/n$ not just that the limit
of the exponent is $0$. Its explained in my response to
Jameson's post.

RonL
• Jan 7th 2006, 03:11 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Jameson's statement is a fallacy - you need to consider the rate of
growth of what is raised to $1/n$ not just that the limit
of the exponent is $0$. Its explained in my response to
Jameson's post.

RonL

I know, I told him that.

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, 09:18 AM
Jameson
Quote:

Originally Posted by CaptainBlack
This is invalid as it pays no attention to the rate of growth of what is
being raised to $1/n$. Such an argument would apply to

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

for any real function $f(n)$.

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

RonL

I don't understand your point. Obviously $lim_{x\rightarrow\infty}n^n=\infty$ I was simply telling her to analytically think about the trend of the exponent. If I made a mistake I apologize, but I really don't see how I was false in showing that as n approaches infinity, the function becomes $n^0$
• Jan 8th 2006, 09:24 AM
Jameson
Quote:

Originally Posted by ThePerfectHacker
Jameson I believe the rule goes that if the composition of the limits exists then the limit is the functional composition. That means what you said about looking at the exponent part lacks mathematical rigor.

Yes it did lack rigor. I like your method better. I make a quick post and I as I said above I was trying to analytically show the trend of the function. I'm still not seeing my fallacy though. :confused:
• Jan 8th 2006, 10:06 AM
CaptainBlack
Quote:

Originally Posted by Jameson
I don't understand your point. Obviously $lim_{x\rightarrow\infty}n^n=\infty$ I was simply telling her to analytically think about the trend of the exponent. If I made a mistake I apologize, but I really don't see how I was false in showing that as n approaches infinity, the function becomes $n^0$

Your argument about the exponent if valid would also prove that:

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

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, 10:38 AM
ThePerfectHacker
Jameson, your problem, informally was that you looked at the exponent you should have also seen the base thus it is of the form $\infty^0$ no conclusion could be drawn.
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