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Mathematics 18 Online
OpenStudy (emanuel992):

Can I learn advanced math without go to university?

OpenStudy (3mar):

Of course.

OpenStudy (steve816):

Anything is possible as long as you are determined to achieve it!

OpenStudy (steve816):

What math do you speak of?

OpenStudy (3mar):

But you have to know what courses they study to take them online or with private instructor or anyway!

OpenStudy (steve816):

You don't even have to take it online or with private instructor, you can just self study any material @3mar

OpenStudy (steve816):

Of course it's difficult, but it is doable.

OpenStudy (3mar):

You should have been taught to know and learn, if not, you have to have a reference in order to solve and facilitate and smooth the difficulties and obstacles you face!

OpenStudy (emanuel992):

I think that if i go to university i can learn faster than if I study in my home, but maybe learn anything is better than learn nothing

OpenStudy (steve816):

Well, of course you need to have perquisites as all of math is in layers, but nowadays, there are many resources online that can help a lot.

OpenStudy (emanuel992):

but sometimes i feel so lonely and I don't know if I am enought smat for do math, and falling down

OpenStudy (emanuel992):

yeah, we are living in the information age

OpenStudy (blazeryder):

Try Khan Academy. They have courses from kindergarten math to College level.

OpenStudy (emanuel992):

if there is a limit in a sequence, then this is the only one. Demostration:

OpenStudy (emanuel992):

l_0 and l_1 are limits of {a_n}, for each e>0, there n_0, n_1 such that |a_n- l_0| <e for n>n_0. And |a_n-l_1|<e fo n>n_1. So n>max(n0,n1): |l_0-l_1|<=|l_0-a_n|+|a_n-l_1|<e+e, it is tru for each e>0, so l_0=l_1 I don't know how to follow the demostrations on my book

OpenStudy (emanuel992):

\[|l_0 - l_1|\] why they do that?

OpenStudy (emanuel992):

thank you so much for ask my questions

OpenStudy (3mar):

You are welcome!

OpenStudy (steve816):

No problem!

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