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?


1 Answer 1


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

  • $\begingroup$ What I don't fully understand about math proofs are the seemingly arbitrary starting points. Is there some reason that we introduce the inequality you started your proof with? Other than we know by eliminating y^4 from the denominator that it makes for easy cancellation in the numerator? $\endgroup$
    – Jim
    Oct 5, 2012 at 15:40
  • $\begingroup$ @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 $\lim_{(x,y)\rightarrow (a,b)}f(x,y)=L$. $\endgroup$
    – blindman
    Oct 5, 2012 at 15:45
  • $\begingroup$ @Jim M: You are welcome. I am ready to help you. $\endgroup$
    – blindman
    Oct 5, 2012 at 15:54

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