# Evaluate $\lim_{(x,y)\to (0,0)}\frac{x^3+y^4}{x^2+y^2}$

$$\lim_{(x,y)\to (0,0)}\frac{x^3+y^4}{x^2+y^2}$$

What I have manage to come up with is:

$$\lim_{(x,y)\to (0,0)}\left|\frac{x^3+y^4}{x^2+y^2}\right|\leq \lim_{(x,y)\to (0,0)}\left|\frac{x^3+y^4}{2xy}\right|\leq \lim_{(x,y)\to (0,0)}\left|\frac{x^4+y^4}{2xy}\right|$$

taking $x=r\cos\theta$ and $y=r\sin\theta$

we get $$\lim_{r\to 0}\left|\frac{r^4(\cos^4\theta+\sin^4\theta)}{2r^2\cos\theta\sin\theta}\right|=\lim_{r\to 0}\left|\frac{r^4}{2}\cdot\frac{\cos^4\theta+\sin^4\theta}{2\cos\theta\sin\theta}\right|$$ which is a bounded function and a function going to zero, and there for the limit is zero and by the squeeze limit

$$\lim_{(x,y)\to (0,0)}\frac{x^3+y^4}{x^2+y^2}=0$$

Is the way is valid? is there a simpler way?

• Just doing polar coordinates right off the bat would probably be easier.
– Dave
Mar 9, 2018 at 15:08
• Please remember that you can choose an aswer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/…
– user
Mar 12, 2018 at 23:42

For both $x,y\ne 0$, then \begin{align*} \left|\frac{x^{3}+y^{4}}{x^{2}+y^{2}}\right|&\leq\frac{|x|^{3}}{|x|^{2}+|y|^{2}}+\frac{|y|^{4}}{|x|^{2}+|y|^{2}}\\ &\leq\frac{|x|^{3}}{|x|^{2}}+\frac{|y|^{4}}{|y|^{2}}\\ &=|x|+|y|^{2}, \end{align*} either $x=0$ or $y=0$ (but not both), the above inequality still holds, so \begin{align*} \left|\frac{x^{3}+y^{4}}{x^{2}+y^{2}}\right|\leq|x|+|y|^{2},~~~~(x,y)\ne(0,0), \end{align*} and we know that $|x|+|y|^{2}\rightarrow 0$ as $(x,y)\rightarrow(0,0)$, so the result follows by Squeeze Theorem.

We can observe directly by polar coordinates

$$\frac{x^3+y^4}{x^2+y^2}=r\cdot f(r,\theta)\to 0$$

since $f(r,\theta)=\cos^3 \theta+r\sin^4 \theta$ is bounded.

There's a problem with the step:

$$\lim_{(x,y)\to (0,0)}\left|\frac{x^3+y^4}{2xy}\right|\leq \lim_{(x,y)\to (0,0)}\left|\frac{x^4+y^4}{2xy}\right|$$

Note that $x^3 \ge x^4$ for $|x| \le 1$.

However, the idea about polar coordinates should work and is even easier if you perform the substitution right away:

$$\lim_{(x,y)\to (0,0)}\frac{x^3+y^4}{x^2+y^2} = \lim_{r \to 0}\frac{r^3 \cos^3 \theta + r^4 \sin^4 \theta }{r^2} = \lim_{r \to 0}r \cos^3 \theta + r^2 \sin^4 \theta.$$

Note that $$|r \cos^3 \theta + r^2 \sin^4 \theta| \le r|\cos\theta|^3 + r^2|\sin \theta|^4 \le r + r^2 \to 0,$$ so the limit is $0$ by squeeze theorem.