# Evaluating the limit of multivariable function

Given a function $$\lim_{(x, y)\to(\infty, \infty)} \frac{x+y+2x^2+2y^2}{x^3+y^3}$$ find the limit or prove that it does not exist. To check whether the limit exists, the direction towards the point should be checked, but how one proceeds if both variable tends to infinity?

Substituting for $$x = r\cos\theta, y = r\sin\theta$$ $$\lim_{r \to\infty}\frac{r\cos\theta + r\sin\theta+2r^2\cos^2\theta + 2r^2\sin^2\theta}{r^3\cos\theta + r^3\sin\theta} = 0.$$

It shows that limit is zero, I am not sure if it's correct way to approach the problem.

• try to use polar coordinates – ALG Nov 25 '18 at 11:39
• @ALG By substitution, I showed that limit is zero, and the result does not depend on the angle. However, for me it's not clear whether the solution is correct. – Inter Veridium Nov 25 '18 at 13:50

The limit doesn't exist indeed let as $$t \to \infty$$

• $$x=t$$
• $$y=-t+\frac1t$$

then

$$\frac{x+y+2x^2+2y^2}{x^3+y^3}=\frac{t-t+\frac1t+2t^2+2\left(t^2-2+\frac1{t^2}\right)}{t^3+\left(-t^3+3t-3\frac1t+\frac1{t^3}\right)}=\frac{4t^2-4+\frac1t+2\frac1{t^2}}{3t+3\frac1t+\frac1{t^3}}\to \infty$$

but for $$x=y$$ it is equal to $$0$$.

• Much appreciated! If the substitution were $1/t$, would limit be then changed to $t\to0$? – Inter Veridium Nov 25 '18 at 19:22
• @InterVeridium You are welcome! Usually when the denominator can goes to zero for values different from the limit point the the limit is very prone to do not exists. here the trick was make $x^3+y^3$vanish bu keeping some smaller term. Yes we can set $x=1/u$ and $y=1/v$ and take the limit as $(u,v)\to (0,0)$ but I'm not sure it can help so much. – gimusi Nov 25 '18 at 19:27