Finding limit of multivariable function.

Question

I have a question that might be silly but I need to understand these kind of problems. So i have this limit:$$\lim_{(x,y)\to(0,0)} \frac{x^3 + y^3}{x^2 + y^2}$$ To solve it I am going to use poolar coordinates, so the limit would be like this:

$$\lim_{r\to0} \frac{r^3\cos^3\theta + r^3\sin^3\theta}{r^2}=$$ $$\lim_{r\to0} {r(\cos^3\theta + \sin^3\theta)}$$

Now it is clear to me that r tends to $r\to0$ but, can I actually say something about $\cos^3\theta + \sin^3\theta?$ In case I could, then I would say that it is bounded, thus the limit would be 0. But it is not clear to me, could anyone help me with this?

Answer 1

It works because

$$\left|\cos^3\theta + \sin^3\theta\right| \le \left|\cos\theta\right|^3 + \left|\sin\theta\right|^3 \le 2$$

Also without polar coordinates:

\begin{align} \left|\frac{x^3+y^3}{x^2+y^2}\right| &= \frac{|x+y||x^2-xy+y^2|}{x^2+y^2} \\ &\le|x+y| \cdot \frac{x^2+y^2+|xy|}{x^2+y^2} \\ &\le \sqrt{2}(x^2+y^2) \cdot \frac{\frac32(x^2+y^2)}{x^2+y^2}\\ &= \frac{3\sqrt{2}}2\|(x,y)\|^2 \xrightarrow{(x,y) \to (0,0)} 0 \end{align}

Answer 2

Both cosine and sine are always bounded by $1$ in absolute value. Therefore, $|\cos^3 \theta + \sin^3 \theta| \leq 2$. Your limit is therefore $0$.

Answer 3

You can write $$ \frac{x^3+y^3}{x^2+y^2}= \frac{x^2+y^2-xy}{x^2+y^2}(x+y)=x+y-\frac{xy}{x^2+y^2}(x+y) $$ Now observe that $$ \left|\frac{xy}{x^2+y^2}\right|\le1 $$ because $a^2-ab+b^2\ge0$.