Hello,

I have got some doubts about sequences....
Are u able to help?

(1) Find the limits of the convergent sequnces (log10 nsqr) /n
(2) Find the limits of the convergent sequnces (1 + 3/n)power n

Where n in eaxh case is a positive integer......

(3) Find the series sum (-1) power n [ Sqr root n/(n +1)]

Use any method to determine it's convergence and divergence.....

(4) Is this differential eqn dy/dx = (3xpower2 ylny) /(2y-xpower3-xpower3 lny)
exact?

I'm also confused when it comes to having ln function......
Any one can help me on this?

 

mmm homework.

why are you confued with ln? can you use log? if so ln is just log to base e and it follows exactly the same rules (since it is a log function) the only thing is the inverse is not 10^x but exp(xP

 

no homework.....
It's exam question.....
Any one can help me?

 

I hope it's not too late

for 1) I'm assuming you mean lim (log(n^2))/n. If this is the case, you can take the log(n^2) and change it to 2*log(n). After that, we see that n > log(n) as n->infinity. Therefore, limit = 0.

2) as n->infinity, 3/n -> 0. Then, 1^n = 1. Thus, limit as n->infinity = 1.

3) I'd use the Alternating Series test. This gives us two criteria to find: is An+1 < An, and is lim An = 0? Both of these are satisfied, so it is convergent.

4) We rearrange this a bit:

-3x^2*y*ln(y) + (2*y - x^3 - x^3*ln y)*(dy/dx) = 0.

An exact equation is in the form P(x) + Q(x)(dy/dx) = 0.

P(x) = -3x^2*y*ln(y) and Q(x) = 2*y - x^3 - x^3*ln(y)

To find out if the equation is exact, the partial derivative of P(x) wrt y must equal the partial of Q(x) wrt x.

Py = -3*x^2*ln(y) - 3*x^2
Qx = -3*x^2-3*x^2*ln(y)

Py = Qx, therefore the equation is exact.

 

Sorry.......
(1) is [log base 10 n^2 /n]
How do i get the limits?
Pls advise.

 

The result posted for #1 is correct and I just want to add the intermediate step that I think Shermaine is asking about... I think...

Log is Log base 10 and not Natural Log.

We agree that Log (n^2) = 2*Log(n)
So the problem is evaluate Lim (n->InF) [2*Log(n) / n]

The numerator tends towards infinity.
The denominator tends towards infinity.

This is indeterminate and we must use L'Hospitals rule by evaluating:

Lim (n->Inf) [D(2Log(n)) / D(n)] where D is the derivative wrt n.

Doing so, D[2*Log(n)] = 1 / (LN(10)*n)
and D[(n)] = 1

now, we take Lim (n->Inf) [2/(LN(10)n)] / 1

This expression tends towards 0 and the rule stated that this limit is equivalent to the original problem so the answer is 0.

 

The result posted for #1 is correct and I just want to add the intermediate step that I think Shermaine is asking about... I think...

Log is Log base 10 and not Natural Log.

We agree that Log (n^2) = 2*Log(n)
So the problem is evaluate Lim (n->InF) [2*Log(n) / n]

The numerator tends towards infinity.
The denominator tends towards infinity.

This is indeterminate and we must use L'Hospitals rule by evaluating:

Lim (n->Inf) [D(2Log(n)) / D(n)] where D is the derivative wrt n.

Doing so, D[2*Log(n)] = 1 / (LN(10)*n)
and D[(n)] = 1

now, we take Lim (n->Inf) [2/(LN(10)n)] / 1

This expression tends towards 0 and the rule stated that this limit is equivalent to the original problem so the answer is 0.

 

Hello,

Can we change to log to ln.??
Something like log base 10 n ^ 2 = ln 10/ ln 2?
How do we do it if were to change it from log to ln?
Pls advise.

 

ln(x) = log(x)/Log(e)

if that is of any help

 

Ahh yes, I always forget that part .