# limits of a sequence

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• May 10th 2010, 09:14 AM
Mathman87
limits of a sequence
In general, how do you go about finding the limit of a sequence?

e.g. n^{1/ $\sqrt{n}$}

Sorry about latex fail, its n to the power the whole of the bracket.
• May 10th 2010, 09:43 AM
dwsmith
Quote:

Originally Posted by Mathman87
In general, how do you go about finding the limit of a sequence?

e.g. n^{1/ $\sqrt{n}$}

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

What is n approaching?
• May 10th 2010, 09:51 AM
Mathman87
It doesnt specify on the examples, but i imagine infinity?
• May 10th 2010, 09:53 AM
dwsmith
$\infty^0$ is an indeterminant form. Do you know how to solve those?
• May 10th 2010, 09:57 AM
Mathman87
no?
• May 10th 2010, 10:01 AM
dwsmith
L'Hopital's Rule
• May 10th 2010, 10:29 AM
AllanCuz
Quote:

Originally Posted by Mathman87
In general, how do you go about finding the limit of a sequence?

e.g. n^{1/ $\sqrt{n}$}

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

$\lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } }$

Let,

$y = n^{ \frac{1}{ \sqrt{n} } }$

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

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

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

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

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

Thus,

$\lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } } = \lim_{ n \to \infty } y = \lim_{ n \to \infty } e^{ln(y)}$ => $e^ { \lim_{ n \to \infty } ln(y) } = e^0 = 1$
• May 10th 2010, 10:33 AM
Mathman87
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?
• May 10th 2010, 10:36 AM
AllanCuz
Quote:

Originally Posted by Mathman87
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).
• May 10th 2010, 10:46 AM
Mathman87
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!
• May 10th 2010, 10:53 AM
AllanCuz
Quote:

Originally Posted by Mathman87
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!

Dude, you need to add limits!

$\frac {log(n)} {n^{1/10} }$

This is an equation and L'Hopitals has nothing to do with it!

However, if we have

$\lim_{ n \to \infty } \frac {log(n)} {n^{1/10} }$

This is of the form $\frac { \infty }{ \infty }$ and we can use L'hopitals.

$\lim_{ n \to \infty } \frac {log(n)} {n^{1/10} } = \lim_{ n \to \infty } \frac{ \frac{1}{n} }{ \frac{1}{10 n^{ \frac{9}{10} } } } = \lim_{ n \to \infty } \frac{ 10 n^{ \frac{9}{10} } }{n}$

$10 \lim_{ n \to \infty } \frac{1}{n^{ \frac{1}{10} } } = 0$
• May 10th 2010, 11:36 AM
Mathman87
I know, but it doesnt tell us what n is tending towards on the question, hence why im a bit confused?
Iv attatched the document so you can see what i mean. Its question 4....
• May 10th 2010, 11:44 AM
dwsmith
Since this is a test from 2007/2008, I would assume the prof. may have stated what the limits are approaching in class.

4
(i) is probably infinity and then the answer is e
• May 10th 2010, 11:48 AM
dwsmith
(iii) proabably 0
$\frac{6n^2+11n+6}{n^2+3n+2}\rightarrow \lim_{n\to 0}\frac{6n^2}{n^2}=6$

(iv)+(v)
AllanCuz answered

(ii) is probably supposed to be approaching infinity