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 . 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 ? And what is any number or variable to that answer?
This happens to be a famous limit
I will prove that =1.
Notice that the function is a monotonic decreasing function, which has a lower bound. Thus, has a limit. Call that thus,
L=
thus,
But because is countinous for that interval,
Thus,
But this fits the necessary condition of L'Hopital's Rule,
Thus, thus,
Finally putting all of this together,
=1
Now:Originally Posted by TexasGirl
,
and as the exponential function is continuous if the limit exists:
,
so we need only worry about:
.
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 , there exists a such that:
,
for all sufficiently large. Which is sufficient to show that
,
for any . Of course this can also be demonstrated
using L'Hopital's rule as is done for elsewhere in this thread.
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 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 becomesOriginally Posted by CaptainBlack