Question regarding limits of multivariable functions - is it really zero? 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?
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: $$\lim_{x\rightarrow a}f(x)=L\iff \forall \epsilon&gt;0\exists \delta&gt;0\forall x((\color{red}{0&lt;}\vert x-a\vert&lt;\delta)\implies (\vert f(x)-L\vert&lt;\epsilon)).$$ Note the "$0&lt;$" 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 $\epsilon&gt;0$, 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 $\vert{0\over x}-0\vert&lt;\epsilon$.



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.
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?