- #1

- 1,796

- 53

## Main Question or Discussion Point

Hi,

Suppose I have

[tex]\lim_{r\to 0} \left\{\int_0^{\pi} \frac{f(r,t)}{r^2}dt - \int_0^{\pi} \frac{g(r,t)}{r} dt\right\}[/tex]

and both integrals tend to infinity. So I combine them:

[tex]\lim_{r\to 0} \int_0^{\pi} \frac{f(r,t)-r g(r,t)}{r^2} dt[/tex]

now at this point, the numerator in the integrand does not go to zero as r goes to zero but rather [itex]\cos(t)[/itex] so it does go to zero at one point in the interval of integration. Can I apply L'Hospital's rule and conclude:

[tex]\lim_{r\to 0} \int_0^{\pi} \frac{f(r,t)-r g(r,t)}{r^2} dt\overset{?}{=}\lim_{r\to 0} \int_0^{\pi} \frac{\frac{d}{dr}\left[f(r,t)-r g(r,t)\right]}{\frac{d}{dr}r^2}dt[/tex]

Ok thanks,

Jack

Suppose I have

[tex]\lim_{r\to 0} \left\{\int_0^{\pi} \frac{f(r,t)}{r^2}dt - \int_0^{\pi} \frac{g(r,t)}{r} dt\right\}[/tex]

and both integrals tend to infinity. So I combine them:

[tex]\lim_{r\to 0} \int_0^{\pi} \frac{f(r,t)-r g(r,t)}{r^2} dt[/tex]

now at this point, the numerator in the integrand does not go to zero as r goes to zero but rather [itex]\cos(t)[/itex] so it does go to zero at one point in the interval of integration. Can I apply L'Hospital's rule and conclude:

[tex]\lim_{r\to 0} \int_0^{\pi} \frac{f(r,t)-r g(r,t)}{r^2} dt\overset{?}{=}\lim_{r\to 0} \int_0^{\pi} \frac{\frac{d}{dr}\left[f(r,t)-r g(r,t)\right]}{\frac{d}{dr}r^2}dt[/tex]

Ok thanks,

Jack