limits of a sequence

Apr 2010
30
0
In general, how do you go about finding the limit of a sequence?

e.g. n^{1/\(\displaystyle \sqrt{n}\)}

Sorry about latex fail, its n to the power the whole of the bracket.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
In general, how do you go about finding the limit of a sequence?

e.g. n^{1/\(\displaystyle \sqrt{n}\)}

Sorry about latex fail, its n to the power the whole of the bracket.
What is n approaching?
 
Apr 2010
30
0
It doesnt specify on the examples, but i imagine infinity?
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
\(\displaystyle \infty^0\) is an indeterminant form. Do you know how to solve those?
 
Apr 2010
384
153
Canada
In general, how do you go about finding the limit of a sequence?

e.g. n^{1/\(\displaystyle \sqrt{n}\)}

Sorry about latex fail, its n to the power the whole of the bracket.
\(\displaystyle \lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } } \)

Let,

\(\displaystyle y = n^{ \frac{1}{ \sqrt{n} } } \)

\(\displaystyle ln(y) = \frac{1}{ \sqrt{n}} ln(n) \)

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

This is of the form \(\displaystyle \frac{ \infty } { \infty } \) so we can use L'hopitals here to get

\(\displaystyle \lim_{ n \to \infty } \frac{1}{ \sqrt{n}} ln(n) = \lim_{ n \to \infty } \frac{ \frac{1}{n} } { \frac{1}{2 \sqrt{n} } } = \lim_{ n \to \infty } \frac{2 \sqrt{n} }{n} = \lim_{ n \to \infty } \frac{2}{ \sqrt{n} } =0 \)

Remember that this was the limit of \(\displaystyle ln(y) \). So what we can do is \(\displaystyle e^{ln(y)} = y \)

Thus,

\(\displaystyle \lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } } = \lim_{ n \to \infty } y = \lim_{ n \to \infty } e^{ln(y)} \) => \(\displaystyle e^ { \lim_{ n \to \infty } ln(y) } = e^0 = 1 \)
 
  • Like
Reactions: dwsmith
Apr 2010
30
0
Ok, so can L'hopitals rule always be used to find the limit of sequences?

Like if i had {(2n+3)(3n+2)}/{(n^2)+3n+2} would the limit be 6?

Also if i had {log(n)}/{10th root of n}..... sorry latex fail again, would i be able to use L'hopitals rule this time?
 
Apr 2010
384
153
Canada
Ok, so can L'hopitals rule always be used to find the limit of sequences?

Like if i had {(2n+3)(3n+2)}/{(n^2)+3n+2} would the limit be 6?

Also if i had {log(n)}/{10th root of n}..... sorry latex fail again, would i be able to use L'hopitals rule this time?
You can use L'Hopitals anytime there is a limit of indeterminate form. And you cannot ask what the limit of a function or a series of functions will be without providing where our limit is taken (i.e. x--> infinity or x-->2).
 
Apr 2010
30
0
Ok i see. So i tried using L'hopitals rule for {log(n)}/{n^(1/10)}

So i differentiated the denominator and the numerator and ended up with 10*{n^(-1/10)}. Is this right?

Im not convinced im using it right lol!