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 $
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?
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} $
$\displaystyle 10 \lim_{ n \to \infty } \frac{1}{n^{ \frac{1}{10} } } = 0$