1. ## 2-variable limit

[x^5 + y^2]/[x^4 + |y|] -> 0 as (x,y) -> (0,0)

2. ## Re: 2-variable limit

Originally Posted by marqushogas
$\displaystyle x = r\cos{\theta}$ and $\displaystyle y = r\sin{\theta}$, so
\displaystyle \begin{align*} \lim_{(x, y) \to (0, 0)}\frac{x^5 + y^2}{x^4 + |y|} &= \lim_{r \to 0}\frac{r^5\cos^5{\theta} + r^2\sin^2{\theta}}{r^4\cos^4{\theta} + |r||\sin{\theta}|} \\ &= \lim_{r \to 0}\frac{r^2\left(r^3\cos^5{\theta} + \sin^2{\theta}\right)}{|r|\left(|r|r^2 \cos^4{\theta} + | \sin{\theta} | \right)} \\ &= \lim_{r \to 0}\frac{|r|\left(r^3\cos^5{\theta} + \sin^2{\theta} \right)}{|r|r^2\cos^5{\theta} + | \sin{\theta} |} \\ &= \frac{0\left( 0 + \sin^2{\theta} \right)}{0 + | \sin{\theta} |} \\ &= \frac{0}{|\sin{\theta}|} \\ &= 0 \end{align*}
$= \frac{0}{|\sin{\theta}|}= 0$
$\frac{0}{|\sin{\theta}|}= 0 \Leftrightarrow \sin \theta\neq 0$ and $\sin \theta =0$ iff $(x,y)$ belongs to the set $S=\{(x,0):x\neq 0\}$ . Now, $f(x,y)=x$ on $S$ and the limit of $f(x,y)$ along $S$ for $(x,y)\to (0,0)$ is also $0$ .