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.

Printable View

- May 10th 2010, 09:14 AMMathman87limits of a sequence
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. - May 10th 2010, 09:43 AMdwsmith
- May 10th 2010, 09:51 AMMathman87
It doesnt specify on the examples, but i imagine infinity?

- May 10th 2010, 09:53 AMdwsmith
$\displaystyle \infty^0$ is an indeterminant form. Do you know how to solve those?

- May 10th 2010, 09:57 AMMathman87
no?

- May 10th 2010, 10:01 AMdwsmith
L'Hopital's Rule

- May 10th 2010, 10:29 AMAllanCuz
$\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 $ - May 10th 2010, 10:33 AMMathman87
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 AMAllanCuz
- May 10th 2010, 10:46 AMMathman87
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 AMAllanCuz
Dude, you need to add limits!

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

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

However, if we have

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

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

$\displaystyle \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} $

$\displaystyle 10 \lim_{ n \to \infty } \frac{1}{n^{ \frac{1}{10} } } = 0$ - May 10th 2010, 11:36 AMMathman87
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 AMdwsmith
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 AMdwsmith
(iii) proabably 0

$\displaystyle \frac{6n^2+11n+6}{n^2+3n+2}\rightarrow \lim_{n\to 0}\frac{6n^2}{n^2}=6$

(iv)+(v)

AllanCuz answered

(ii) is probably supposed to be approaching infinity