Now we get to actual calculus dealing with functions of two variables, and we naturally begin with limits. Recall from Calculus 1 the definition of a limit:

A function \(f(x)\) has limit \(L\) as \(x\) approaches \(c\) if \(|f(x)-L|\) (the distance between the function values and the limit) gets arbitrary small (\(|f(x)-L| < \epsilon\)) for values of \(x\) that are sufficiently close to \(c\) (\(|x-c| < \delta\)).

Let's extend this definition to functions of two variables in \(\mathbb{R}^3\). Instead of \(x\) approaching \(c\), we'll have \((x,y)\) approaching \((a,b)\). So here's the definition of what it means for \[\lim_{(x,y) \to (a,b)} f(x,y) = L:\]

A function \(f(x,y)\) has limit \(L\) as \((x,y) \to (a,b)\) if \(|f(x,y) - L|\) can be made arbitrarily small for points sufficiently close to \((a,b)\).

The rules for limits with functions of two variables are essentially identical to the ones for limits with functions of one variable in Calculus 1.

First, suppose that \(\lim_{(x,y) \to (a,b)} f(x,y) = L\) and \(\lim_{(x,y) \to (a,b)} g(x,y) = M\).

Evaluate \(\displaystyle\lim_{(x,y) \to (5,-2)} x^5+4x^3y-5xy^2\).

Use the limit rules outlined above to break this down to basic forms:

\[\begin{align} \lim_{(x,y) \to (5,-2)} x^5+4x^3y-5xy^2 &= \lim_{(x,y) \to (5,-2)} x^5 + \lim_{(x,y) \to (5,-2)} 4x^3y - \lim_{(x,y) \to (5,-2)} 5xy^2\\ &= \left(\lim_{(x,y) \to (5,-2)} x\right)^5 + 4\left(\lim_{(x,y) \to (5,-2)} x\right)^3\left(\lim_{(x,y) \to (5,-2)} y\right)\\ &- 5\left(\lim_{(x,y) \to (5,-2)} x\right)\left(\lim_{(x,y) \to (5,-2)} y\right)^2\\ &= 5^5+4(5)^3(-2)-5(5)(-2)^2\\ &= \ans{2025} \end{align}\]We could evaluate this limit by simply evaluating the function at \((5,-2)\) because the function is continuous (a polynomial in this case).

Evaluate \(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{x^2}{x^2+y^2}\).

If we try evaluating the function at \((0,0)\), we'll get 0 in both the numerator and denominator. It turns out that this limit does not exist. How can we prove this? Think about the following picture in two dimensions:

If we approach from one side, the function values approach one value, and if we approach from another side, the function values approach a different value, which means that the two-sided limit doesn't exist.

We can do something similiar in three dimensions: if we approach \((0,0)\) along two different lines and show that the limits are different in each case.

For this one, we'll approach along the x-axis and then the y-axis.

- Along the x-axis: let \(y=0\) \[\lim_{x \to 0} \dfrac{x^2}{x^2+0^2} = \lim_{x \to 0} \dfrac{x^2}{x^2} = 1\]
- Along the y-axis: let \(x=0\) \[\lim_{y \to 0} \dfrac{0^2}{0^2+y^2} = \lim_{y \to 0} 0 = 0\]

Since these two limits differ, we've proven that the limit does not exist. We can graph this surface to observe this behavior. Notice how approaching the origin along the x-axis puts you up on the ridge, where the function values are 1, and approaching along the y-axis puts you in the valley, where the function values are 0.

Evaluate \(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{x^2 \sin^2 y}{x^2+2y^2}\).

For this one, we'll use the Squeeze Theorem, noting that since \(y^2 \geq 0\) regardless of what \(y\) is, that means that \[\dfrac{x^2}{x^2+2y^2} \leq 1 \longrightarrow \dfrac{x^2 \sin^2 y}{x^2+2y^2} \leq \sin^2 y\] Therefore,

\[\begin{align} \lim_{(x,y) \to (0,0)} 0 &\leq \lim_{(x,y) \to (0,0)} \dfrac{x^2 \sin^2 y}{x^2+2y^2} \leq \lim_{(x,y) \to (0,0)} \sin^2 y\\ 0 &\leq \lim_{(x,y) \to (0,0)} \dfrac{x^2 \sin^2 y}{x^2+2y^2} \leq 0\\ &\implies \ans{\lim_{(x,y) \to (0,0)} \dfrac{x^2 \sin^2 y}{x^2+2y^2}=0} \end{align}\]- Evaluate \(\displaystyle\lim_{(x,y) \to (2,9)} 101\) or show that the limit does not exist.
- Evaluate \(\displaystyle\lim_{(x,y) \to (-3,3)} 4x^2-y^2\) or show that the limit does not exist.
- Evaluate \(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{x+2y}{x-2y}\) or show that the limit does not exist.
- Evaluate \(\displaystyle\lim_{(x,y) \to (0,\pi)} \dfrac{\cos xy + \sin xy}{2y}\) or show that the limit does not exist.
- Evaluate \(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{y^4-2x^2}{y^4+x^3}\) or show that the limit does not exist.

\(\displaystyle\lim_{(x,y) \to (2,9)} 101 = 101\)

\(\displaystyle\lim_{(x,y) \to (-3,3)} 4x^2-y^2 = 27\)

\(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{x+2y}{x-2y} \ DNE\)

\(\displaystyle\lim_{(x,y) \to (0,\pi)} \dfrac{\cos xy + \sin xy}{2y} = \dfrac{1}{2\pi}\)

\(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{y^4-2x^2}{y^4+x^3} \ DNE\)

In general, if approaching the origin along the axes does not work/apply, try substituting \(y=mx+b\).

Show that \(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{4xy}{3x^2+y^2}\) does not exist.

Since this is different for different values of \(m\), we conclude that this limit does not exist.

Show that \(\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{y}{\sqrt{x^2-y^2}}\) does not exist.

(work it out)

Recall from Calculus 1 that we define continuity using limits; specifically, if the limit exists (approaching from both sides gives the same result) and equals the function value at that point, the function is continuous there. Therefore, there are three conditions (the limit exists, the function value is defined, and the two are equal):

A function \(f\) is continuous at \(c\) if \(\displaystyle\lim_{x \to c} f(x) = f(c)\).

The same is true with functions of two variables: a function \(f(x,y)\) is continuous at a point \((a,b)\) if \[\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b).\]

Is the following function continuous?

\[f(x,y) = \left\{\begin{array}{c l} \dfrac{3xy^2}{x^2+y^4} & \textrm{ if } (x,y) \neq (0,0)\\ 0 & \textrm{ if } (x,y) = (0,0) \end{array}\right.\]The limit of this function as \((x,y)\) approaches \((0,0)\) does not exist. To prove this, we need to substitute \(x=my^2\) (a linear substitution isn't good enough):

\[\lim_{(x,y) \to (0,0)} \dfrac{3xy^2}{x^2+y^4} = \lim_{x \to 0} \dfrac{3my^4}{(m^2+1)y^4} = \dfrac{3m}{m^2+1}\]Since this limit does not exist, the function is not continuous.

At what points of \(\mathbb{R}^2\) is \(f(x,y)=x^2-6x+7xy^2\) continuous?

Since this is a polynomial, it is continuous at all points in \(\mathbb{R}^2\). There's no issue with dividing by 0 or taking the square root of a negative number.

- At what points of \(\mathbb{R}^2\) is \(f(x,y)=\dfrac{xy}{x^2y^2+1}\) continuous?
- At what points of \(\mathbb{R}^2\) is \(f(x,y)=\dfrac{2xy}{x^2-y^2}\) continuous?
- At what points of \(\mathbb{R}^2\) is \(f(x,y)=\dfrac{x^2+y^2}{x(y^2-1)}\) continuous?

All points in \(\mathbb{R}^2\)

\(\mathbb{R}^2 \setminus \{(x,y)\ :\ y^2=x^2\}\)

\(\mathbb{R}^2 \setminus \{(x,y)\ :\ x=0 \textrm{ or } y=1, -1\}\)

If we compose two continuous functions, we'll get another continuous function: \(f\big(g(x,y)\big)\) is continuous at \((a,b)\) if \(g\) is continuous at \((a,b)\) and \(f\) is continuous at \(g(a,b)\).

At what points of \(\mathbb{R}^2\) is \(f(x,y)=\sqrt{x^2+y^2}\) continuous?

Since \(x^2+y^2\) is continuous for all values of \(x\) and \(y\), and \(\sqrt{x}\) is continuous as long as \(x \geq 0\), this compositive function is continuous on the following set:

\[\{(x,y) \in \mathbb{R}^2 \ :\ x^2+y^2 \geq 0\}\]- At what points of \(\mathbb{R}^2\) is \(f(x,y)=\sin xy\) continuous?
- At what points of \(\mathbb{R}^2\) is \(f(x,y)=\sqrt{4-x^2-y^2}\) continuous?
- At what points of \(\mathbb{R}^2\) is \(f(x,y)=\ln(x^2+y^2)\) continuous?

All points in \(\mathbb{R}^2\)

\(\{(x,y) \in \mathbb{R}^2 \ :\ x^2+y^2 \leq 4\}\)

\(\{(x,y) \in \mathbb{R}^2 \ :\ x^2+y^2 > 0\}\)