Recent content by eibon

Homework Statement Use the transformation x= \sqrt{v- u}, y = u + v to evaluate the double integral of f(x, y) = \frac{x}{(x^2 + y)} over the smaller region bounded by y = x^2, y = 4 − x^2, x = 1. Homework Equations The Attempt at a Solution d:={ (x,y)| -\sqrt{2}<x<1 , x^2<y<...

o wait i miss understood what your last post meant(im very bad at english ) so just ignore my last post, so there quotient is {Bn} and lim{Cn) /lim{An} which is a contradiction? because {bn} diverges

the original question was if An converges to L and L does not equal zero and Bn diverges, then does {An*Bn} diverge, so i think Cn was defined

that the quotient converges

please explain how you would do the contradiction

is this right using epsilons?

ok thanks , just out of curiosity how would you prove it with epsilons

then is this right ? and that says DNE not one in the picture

for the quotient of limits you need both An and Bn to converge to use that, but Bn does not converge so i can't use that. and i don't understand how to get the upper and lower bounds, can you please explain it?

so some thing like this? and for the quotient of limits you need both An and Bn to converge to use that, but Bn does not converge so i can't use that

wait how do you get to |Bn-M/L|<e?

L<An<L +\epsilon \leftrightarrow \frac{1}{L +epsilon} <\frac{1}{An} < \frac{1}{L} and M<AnBn<M+ε then i get \frac{M}{L+epsilon} <Bn< \frac{M+epsilon}{L} now what do i do?

yeah i thought of proof by contradiction to day, but i end up getting (M/(L+\epsilon1)) < Bn< (m+\epsilon2)/L and how do you get it converges to M/L? do i need to let epsilon = something?

i don't see how that helps

Homework Statement if {An} converges and lim {An}=L L\neq 0 and {Bn} diverges then does {An X Nn} diverge? prove formally Homework Equations The Attempt at a Solution can anyone give me a hint or show a solution?