- #1
pbialos
Hi, Today i had a test, and i was wondering if what i did is correct:
I had to tell if the [tex]:lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}[/tex] exists. What i did was to say [tex]lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}=lim_{(x,y) \rightarrow (0,0)} \frac {x} {|x|+|y|}*y=0[/tex] because it is bounded multiplied by y that tends to 0.
Is what i did correct?Something tells me it is not because it was too easy.
The second doubt i have is about the convergence of the Integral:
[tex]\int_{0}^{\infty}\frac {sin(x)} {cos(x)+x^2}[/tex]
My doubt came since the integrand of the series is not always positive. Can i just calculate the convergence of:
[tex]\int_{0}^{\infty}|\frac {sin(x)} {cos(x)+x^2}|[/tex] and if it converges, then the original integral also converges?
I would check the convergence of the second integral dividing it between 0-1 and 1-infinity. The integral on the first interval(between 0 and 1) would converge because the integrand is bounded and continuous for x between 0 and 1, and the integral of the second interval(between 1 and infinity) would also converge by comparison with [tex]\int_{1}^{\infty}\frac {1} {x^2-1}[/tex].
I would really appreciate any help.
Regards, Paul.
I had to tell if the [tex]:lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}[/tex] exists. What i did was to say [tex]lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}=lim_{(x,y) \rightarrow (0,0)} \frac {x} {|x|+|y|}*y=0[/tex] because it is bounded multiplied by y that tends to 0.
Is what i did correct?Something tells me it is not because it was too easy.
The second doubt i have is about the convergence of the Integral:
[tex]\int_{0}^{\infty}\frac {sin(x)} {cos(x)+x^2}[/tex]
My doubt came since the integrand of the series is not always positive. Can i just calculate the convergence of:
[tex]\int_{0}^{\infty}|\frac {sin(x)} {cos(x)+x^2}|[/tex] and if it converges, then the original integral also converges?
I would check the convergence of the second integral dividing it between 0-1 and 1-infinity. The integral on the first interval(between 0 and 1) would converge because the integrand is bounded and continuous for x between 0 and 1, and the integral of the second interval(between 1 and infinity) would also converge by comparison with [tex]\int_{1}^{\infty}\frac {1} {x^2-1}[/tex].
I would really appreciate any help.
Regards, Paul.