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??
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?
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
This is invalid as it pays no attention to the rate of growth of what isOriginally 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
Now: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
Jameson's statement is a fallacy - you need to consider the rate ofOriginally 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
I know, I told him that.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.
I don't understand your point. Obviously $\displaystyle lim_{x\rightarrow\infty}n^n=\infty$ 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 $\displaystyle n^0$Originally Posted by CaptainBlack
Your argument about the exponent if valid would also prove that:Originally Posted by Jameson
$\displaystyle \lim_{x\rightarrow\infty}(n^n)^{1/n}=1$,
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