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,

\(\displaystyle \lim_{ n \to \infty } n^{ \frac{1}{ \sqrt{n} } } = \lim_{ n \to \infty } y = \lim_{ n \to \infty } e^{ln(y)} \) => \(\displaystyle e^ { \lim_{ n \to \infty } ln(y) } = e^0 = 1 \)