Recent content by mjjoga

thanks, I finally figured it out! mjjoga

a) Suppose that the limit as n goes to infinity sn=0. If (tn) is a bounded sequence, prove that lim(sntn)=0. So I need to show that abs(sntn)<epsilon, and I know that abs(sn)<epsilon. I mean, I know abs(sntn)=abs(sn)abs(tn that didn't help. I don't know how to go about this. I've tried the...

I figured it out. If I multiply by -1/-1 then with the different signs, I can get rid of the addition and subtraction. I did this before I even factored. then it simplifies to 1/n, so I let N>1/epsilon and it works.

hm, but epsilon could be a tiny fraction and make 1/epsilon really big. There must be some sort of trick we can use to get it less than epsilon.

Define N>ε+2. Then whenever n>max(N,2), we have |(n+2)/(n^2-3)|<(n+2)/(n^2-3)<(n+2)/(n^2-4)=1/(n-2)<1/(ε+2-2)=1/ε 1/epsilon isn't less than epsilon. I don't know what I'm doing wrong. i added that n>2 so that the denumerator does not equal 0. I know that 1/(n-2) is right, but what else can i...

so I have 1/(n-2). I have that n>max(epsilon+2,1). I need to get 1/(n-2) < epsilon. I know that 1/(n-2)<1/(epsilon+2-2)=1/epsilon. but 1/epsilon is not always less than epsilon. can you see any errors?

n is in natural numbers

you know what, when I was canceling out the n+2 on the top, I had put down (n+2)[SUP]2 in the denominator (I tried to use the x^2 button) so I left x+2 on the bottom, but it's the sum and difference so I need n-2 on the bottom. then it's fine. Sometimes my algebra just flies out the window...

lim((n+2)/(n^2-3))=0 as n goes to infinity. I can only use the definition of a limit. My work so far, I'm trying to work out what n will be greater than. I have: |(n+2)/(n^2-3)|=(n+2)/(n^2-3) if n is greater than one. From here, I have been trying anything to get rid of the addition...

Yes, it is to infinity, my book wrote it just like that though, it's weird. I was just worried about the abs. value messing it up, but even with it it will be less than or equal to 1. Thanks guys, I appreciate it. mjjoga

ok, your first step has an error in it. "v^2cos^2(x)" should be over 4 because it was over 2 in the original equation. You did the same thing with the second term of the original equation. Changing that will let you add the fractions more easily. Also when you squared 2v-vsinx you should get...

lim (sin⁡(n)/n)=0. The instruction say I can only use the definition of limit and no additional theorems. So the first thing I should do is figure out if l sin(n)/n l < epsilon, find out what n is greater than. I can pull the 1/n out of the absolute value, but I don't know how to get the sine...

Thank you! that makes a lot more sense to me now. mjjoga

ok so, a) If s sub n→0, then for every ε>0 there exists N∈ℝ such that n>N implies s sub n<ε. This a true or false problem. Now this looks like a basic definition of a limit because s sub n -0=s sub n which is less than epsilon. n is in the natural numbers. But, I thought there should be...