limits of a sequence

Mathman87

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
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?

Mathman87

It doesnt specify on the examples, but i imagine infinity?

dwsmith

MHF Hall of Honor
$$\displaystyle \infty^0$$ is an indeterminant form. Do you know how to solve those?

no?

dwsmith

MHF Hall of Honor
L'Hopital's Rule

AllanCuz

AllanCuz

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,

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

dwsmith

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?

AllanCuz

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).

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!