Question regarding limits of multivariable functions - is it really zero?

Question

I am given the following problem in a textbook, and it's a solved problem.

Find $\displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4}$ or justify its nonexistence.

The author justifies the nonexistence of the limit in the following way: he uses the paths $x = 0$ then $y = 0$ and then, to justify its nonexistence, $y = x$. These are the author's notes:

Since

\begin{align} \lim_{(x,0) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4} = \lim_{(x,0) \to (0,0)} \frac{x^2 \cdot 0^2}{x^4 + 3 0^4} = \lim_{(x,0) \to (0,0)} \frac{0}{x^4} = 0 \end{align}

and

\begin{align} \lim_{(0,y) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4} = \lim_{(0,y) \to (0,0)} \frac{0^2 \cdot y^2}{0^4 + 3 y^4} = \lim_{(0,y) \to (0,0)} \frac{0}{3y^4} = 0 \end{align}

but

\begin{align} \lim_{(x,x) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4} = \lim_{(x,x) \to (0,0)} \frac{x^2 \cdot x^2}{x^4 + 3 x^4} = \lim_{(x,x) \to (0,0)} \frac{x^4}{4x^4} = \frac{1}{4} \end{align}

then the limit does not exist.

So my question is: are these limits

\begin{align} \lim_{(0,y) \to (0,0)} \frac{0}{3y^4} \qquad \lim_{(x,0) \to (0,0)} \frac{0}{x^4} \end{align}

really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?

Answer 1

No, both those limits are indeed defined and equal to zero.

The multivariable context is adding unnecessary "noise" here. Consider the simpler expression $$\lim_{x\rightarrow 0}{0\over x},$$ which already captures what's going on.

Remember the precise definition of a limit. Colloquially, we often say something like "$\lim_{x\rightarrow a}f(x)=L$ iff I can get $f$ to be as close as I want to $L$ just by requiring that $x$ be sufficiently close to $a$." However, this misses a very important point which is explicit in the precise formal definition: $$\lim_{x\rightarrow a}f(x)=L\iff \forall \epsilon>0\exists \delta>0\forall x((\color{red}{0<}\vert x-a\vert<\delta)\implies (\vert f(x)-L\vert<\epsilon)).$$ Note the "$0<$" clause in the left hand side: what this says is that we ignore the specific behavior of $f$ at exactly $a$.

This is why $\lim_{x\rightarrow 0}{0\over x}=0$, despite the bad behavior when $x$ is exactly $0$: for any $\epsilon>0$, let $\delta=17$ (say); then it's easy to check that for any $x$ within $17$ of $0$ but not actually equal to $0$, we have $\vert{0\over x}-0\vert<\epsilon$.

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A reasonable question at this point is why we define limits this way. Ultimately that's a question that deserves a serious answer, but very briefly the point is that we're interested in what $f$ "ought to be" at $a$, not what it literally is. Both the derivative and the integral give great examples of this: in each case, we really want to divide by zero but that gives nonsense, so instead we look at what happens when the change in $x$ or the width of our rectangles gets really close to zero without actually being zero. This also gives a nice spin on continuity: intuitively, a function is continuous if it always is what it ought to be.

Answer 2

Both limit

$$\begin{align} \lim_{(0,y) \to (0,0)} \frac{0}{3y^4}=0 \qquad \lim_{(x,0) \to (0,0)} \frac{0}{x^4}=0 \end{align}$$

are correct since the function $f(x,y)$ is identically zero along the trajectories with either $x=0$ or $y=0$.

In other words, what we cannot consider is $x=y=0$ at the same time but all the trajectories with either $x=0$ or $y=0$ are admissible and consistent with the definition of limit.

To clarify better that point see also the related Why are we allowed to cancel fractions in limits?

Answer 3

Indeed, $\lim_\limits{(0,y)\to (0,0)} \frac {0}{3y^4} = 0$ since for all $y$ in the neighborhood of $0, f(0,y) < \epsilon$

However that is not sufficient for $\lim_\limits{(x,y)\to (0,0)} \frac {x^2y^2}{x^4+3y^4}$ . Because $\lim_\limits{(y,y)\to (0,0)} = \frac {y^4}{4y^4} = \frac {1}{4}.$

If the limit exists, it must be have the same value (be within $\epsilon$ of it at least) for all $x,y$ in the neighborhood.