# Proving the limit - multiple variable differentiation

I'm working through an advanced calculus book and want to be certain I understand the idea behind proving limits. This is not homework, I'm just a statistician looking to learn more about mathematics.

The exercise I'm concerned with proving is as follows:

\begin{aligned} \lim_{(x,y)→(0,0)} \frac{x^3y}{x^2 + y^4} \\\ \end{aligned}

My understanding is that I can choose a value to substitute in for y that allows for some easy cancellation that proves the limit equals 0. For instance:

\begin{aligned} x= y^2 \ ; \frac{(y^2)^3y}{(y^2)^2 + y^4} \\\ \end{aligned}

From here, we have:

\begin{aligned} \frac{y^7}{y^4(1 + 1)} \\\ \end{aligned}

Then as y→0 this simplifies to:

\begin{aligned} \frac{0^3}{2} = 0 \\\ \end{aligned}

Is this how the limit could/would be proved?

We observe that $$0\leq\left|\frac{x^3y}{x^2+y^4}\right|\leq \frac{|x^3y|}{x^2}=|xy|$$ for all $x,y\ne 0$. Since $\displaystyle\lim_{(x,y)\rightarrow (0,0)}|xy|=0$ then $$\lim_{(x,y)\rightarrow (0,0)}\frac{x^3y}{x^2+y^4}=0.$$
• @Jim M: We used two results: 1. If $\lim_{(x,y)\rightarrow (a,b)}|f(x,y)|=0$ then $\lim_{(x,y)\rightarrow (a,b)}f(x,y)=0$. 2. If $g(x,y)\leq f(x,y)\leq h(x,y)$ for all $(x,y)$ in a neighborhood of $(a,b)$ and $\lim_{(x,y)\rightarrow (a,b)}g(x,y)=\lim_{(x,y)\rightarrow (a,b)}h(x,y)=L$ then $\lim_{(x,y)\rightarrow (a,b)}f(x,y)=L$. Oct 5 '12 at 15:45