- #26

- 21

- 2

thanks man, this should do the trickYes, but I had the inequalities going the wrong way before.

For small ##h > 0## we have

$$\ln (1+h) > h^2 - \frac 1 2 h^2$$

Apply this to the numerator ##\ln(1 + 2x^2+y^2) > \ln(1 + x^2+y^2) = \ln(1+r^2),## where ##r^2 = x^2+y^2.##

In the denominator we have ##(x^2 + 3 y^2)^2 < (3 x^2 + 3 y^2)^2 = 9 r^4,##

so the function satisfies

$$f = \frac{\ln(1+2x^2 + y^2)}{(x^2 + 3 y^2)^2} > \frac{r^2 - \frac 1 2 r^4}{9 r^4}= \frac 1 9 \left(\frac{1}{r^2} - \frac{1}{2} \right)$$

still think it doesnt include linear paths, or the theta would be a variable not a constant, and in that case you couldnt treat it anymore as a one variable limit and you have the same problem as before. btw, theta still appears in the function i was trying to find the limit.Converting to polar form for this function doesn't help, because the polar form results in different values for different values of the angle. Converting to polar form is helpful in cases where ##\theta## is no longer present.