Recent content by flipsvibe

Ok, so can I say that C=a(n) + b(n) \Rightarrow Sup C \geq a(n) + b(n) \forall n\inN?

Homework Statement Let a(n) and b(n), n\inN, be some real numbers with absolute value at most 1000. Let A={a(n), n\inN}, B={b(n), n\inN}, C={a(n) + b(n), n\inN}. Show that inf A + sup B \leq sup C \leq sup A + sup B The Attempt at a Solution I was thinking that I could show that inf A + sup...

Wait, I think I got this. I didn't write out all of my steps in my last reply..so I'll go back one or two, to one I skipped. (-3/2)[tan(u) / √sec^2(u)] ==>(-3/2) [tanu / √(tan^2u + 1)] ==>-3x / 2√(x^2 + 1)

I plugged in the x = tan u and dx = sec^2 u du, and I got ∫[tan^2(u) + 1]^(-5/2) * sec^2(u) du ==> ∫[sec^2(u)]^(-5/2) * sec^2(u) du ==> ∫[sec^2(u)]^(-3/2) du after integration I end up with (-3/2)(tanu/secu) If I continue on and cancel out the cosines, I end up with (-3/2)sinu. I don't...

So dx = sec^2 u du?

Homework Statement Compute the indefinite integral. ∫(x^2 + 1)^(-5/2) dx The Attempt at a Solution I have a hunch that I need to substitute x = tan(u) but, as always, my lack of trig skills are holding me back.

No progress on the first one at all. I might just try to use L'hopital's rule anyways. We stopped for the semester one chapter before the book goes over it.

That's the way I learned them in high school: derivatives -> integrals. But we are using apostol's calculus, and he goes in the order they were discovered historically.

Thanks, I figured out the first one. I factored out an x^2 in the numerator. Giving me √[x^2 (1+x)] / x(1+2x) ==> x √(1+x) / x(1+2x) ==> lim x --> 0 is equal to 1

Homework Statement Compute the limits. If they don't exist, then explain. 1) lim as x approaches 0 [cos(x) - 1] / [sin^2(x) + x^3] 2) lim as x approaches 0 √(x^2 + x^3) / (x + 2x^2)Homework Equations The Attempt at a Solution 1) I replaced the sin^2x in the denominator with 1 - cos^2x and...