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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 3.2 Multivariable Limits

Subsection 3.2.1 Definition of Limits of Multivariable Functions

I want to re-establish the tools of calculus for multi-variable functions. As with single-variable functions, the starting point is limits. Again, as with single-variable functions, limit are required for the definition of both the deriavtive and the integral. I’ll start with the full, formal, \(\epsilon-\delta\) definition.

Definition 3.2.1.

Let \(f: \RR^n \rightarrow \RR\) be a scalar field. Let \((a_1, a_2, \ldots, a_n)\) be a point in \(\RR^n\text{.}\) Then the statement
\begin{equation*} \lim_{(x_1, x_2, \ldots, x_n) \rightarrow (a_1, a_2, \ldots, a_n)} f(x_1, x_2, \ldots, x_n) = L \end{equation*}
means that \(\forall \epsilon > 0\) \(\exists \delta > 0\) such that if \(|(x_1, x_2, \ldots, x_n) - (a_1, a_2, \ldots, a_n)| \lt \delta\) then \(|f(x_1, x_2, \ldots, x_n) - L| \lt \epsilon\text{.}\)
The definition is essentially the same as the single-variable definition: as the input gets closer and closer to a specific point, the output gets closer and closer to a fixed value \(L\text{.}\) The only issue is that ‘closer and closer’ now happens in \(\RR^n\) instead of \(\RR\text{.}\)
In \(\RR\text{,}\) there were only two directions of approach: from the left and from the right. If the behaviour from both sides was the same, then the limit existed. In \(\RR^n\) for \(n \geq 2\text{,}\) there are infinitely many ways to approach any given point. I can approach along lines in infinitely many directions out from the point. Even more, I can approach along other paths, such as spiral paths or strange jagged paths. This makes it much more difficult to determine the behaviour and much more difficult to prove existence of various limits. However, I do have some good news. First, the definition of continuity remains the same.

Definition 3.2.2.

A scalar field \(f: \RR^n \rightarrow \RR\) is called continuous at \((a_1, a_2, \ldots, a_n)\) if the limit approaching \((a_1,a_2, \ldots, a_n)\) exists and is the same as \(f(a_1,a_2, \ldots, a_n)\text{.}\)
So limits for continuous functions are still reasonable: I just evaluate. But what functions are continuous?
This is the first application of a very important technique: treating the function as a function of only one variable and ignoring the others, pretending that they are constant. If I do that, I end up consider \(n\) different single variable functions. The proposition says that the function is continuous in the new definition if and only if all of the \(n\) different single variable functions are continuous.
This proposition tells me how to recognize continuous functions. Anything involving polynomials, roots, rational functions, trig, exponentials and logarithms is continuous on its domain.

Subsection 3.2.2 Evaluating Multivariable Limits

As I mentioned before, proving the existence of limits is difficult. However, the algebraic techniques of first year calculus can still work for calculations. Here are some examples where I can use algebraic manipulation to simplify limits so that they can be evaluated.
\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{x^2 - y^2}{x-y} = \lim_{(x,y) \rightarrow (0,0)} \frac{(x-y)(x+y)}{(x-y)} \end{equation*}
The limit cannot be directly evaluated. However, with the factoring of the numerator, I can cancel of the \((x-y)\) term. That removes the division by zero problem and lets me evaluate the limit.
\begin{equation*} = \lim_{(x,y) \rightarrow (0,0)} x+y = 0 + 0 = 0 \end{equation*}
This limit means that no matter which direction I choose to approach \((0,0)\) from, the function gets closer and closer to zero.
\begin{equation*} \lim_{(x,y) \rightarrow (4,1)} \frac{ xy - 4y^2}{\sqrt{x} - 2\sqrt{y}} = \lim_{(x,y) \rightarrow (4,1)} \frac{(xy - 4y^2)(\sqrt{x} + 2\sqrt{y})}{(\sqrt{x} - 2 \sqrt{y})(\sqrt{x} + 2 \sqrt{y})} \end{equation*}
This is another limit that cannot be directly evaluated, since it has type \(\frac{0}{0}\text{.}\) To deal with this denominator, I will multiply by its conjugate (in both numerator and denominator, of course). The conjugate gets rid of the square roots in the denominator
\begin{equation*} = \lim_{(x,y) \rightarrow (4,1)} \frac{(xy - 4y^2)(\sqrt{x} + 2\sqrt{y})}{x - 4y} \end{equation*}
Then I can factor a \(y\) out of the numerator and find a common term, \((x-4y)\text{,}\) which I can cancel off.
\begin{equation*} = \lim_{(x,y) \rightarrow (4,1)} \frac{y(x - 4y)(\sqrt{x} + 2\sqrt{y})}{x - 4y} = \lim_{(x,y) \rightarrow (4,1)} y(\sqrt{x} + 2\sqrt{y}) \end{equation*}
At the end, having removed the division by zero, I just evaluate the limit.
\begin{equation*} = \lim_{(x,y) \rightarrow (4,1)} y(\sqrt{x} + 2\sqrt{y}) = (1)(\sqrt{4} + 2 \sqrt{1}) = 2 + 2 = 4 \end{equation*}
\begin{equation*} \lim_{(x,y) \rightarrow (1,2)} \frac{\sqrt{y} - \sqrt{x+1}}{y-x-1} = \lim_{(x,y) \rightarrow (1,2)} \frac{(\sqrt{y} - \sqrt{x+1})(\sqrt{y} + \sqrt{x+1})}{(y - x - 1)(\sqrt{y} + \sqrt{x+1})} \end{equation*}
This is another limit where multiplying by a conjugate is helpful. The conjugate removes the square roots from the numerator.
\begin{equation*} = \lim_{(x,y) \rightarrow (1,2)} \frac{(y-x-1)}{(y-x-1)(\sqrt{y} + \sqrt{x+1})} \end{equation*}
Then I can cancel off the common term, which removes the division by zero. After that, I can simply evaluate the limit.
\begin{equation*} = \lim_{(x,y) \rightarrow (1,2)} \frac{1}{\sqrt{y} + \sqrt{x+1}} = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}} \end{equation*}
When a limit doesn’t exist, I can prove its non-existence by looking at various directions of approach. If the limit along different directions is different, then there can be no cohesive limit. There are several ways to do this, depending on what kind of approach I want to take. I’ll do two examples with different paths of approach, each chosen to suit the specific function under consideration.
Figure 3.2.7. The graph of \(f(x) = \frac{(x+y)^2}{x^2+y^2}\)
\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{(x+y)^2}{x^2 + y^2} \end{equation*}
I will approach \((0,0)\) along the line \(y = mx\text{.}\) That lets me replace \(y\) with \(mx\) in the calculation and the limit because \(x \rightarrow 0\text{.}\)
\begin{align*} \lim_{x \rightarrow 0} \frac{(x+mx)^2}{x^2 + m^2x^2} \amp = \lim_{x \rightarrow 0} \frac{x^2 + 2mx^2 + m^2 x^2}{x^2 + m^2 x^2}\\ \amp = \lim_{x \rightarrow 0} \frac{x^2(1+m)^2}{x^2(1+m^2)} = \frac{(1+m)^2}{1+m^2} \end{align*}
This limit depend on the choice of \(m\text{.}\) I can get infinitely many values (all between 0 and 1) out of this limit depending on which line I use to approach \((0,0)\text{.}\) With all these possible answers, the limit cannot exist. It is interesting to try to visualize the graph: as the values get close to zero, there are pieces of the graph getting close to any number between \(0\) and \(1\text{.}\) Figure 3.2.7 shows some of this behaviour: approaching from the front of figure leads to \(0\text{,}\) but approaching from the sides leads to larger numbers. (The graph is slightly flawed, due to the graphing algorithm. The two cliffs should meet, even though the graph shows a gap between them. Where the two cliffs meet is the line with all the problematic limits.)
In this example, I approach along parabolic paths of the form \(x = m y^2\text{.}\) This lets me replace \(x\) with \(my^2\) and get a single variable limit. Like the previous example, I get a limit that depends on the path of approach, so I conclude that the limit cannot exists.
\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{3xy^2}{x^2 + y^4} = \lim_{y \rightarrow 0} \frac{3my^4}{m^2 y^4 + y^4} = \frac{3m}{m^2+1} \end{equation*}
It is important to note that this technique (choosing multiple paths of approach) can only be used to proves that limits fail. It cannot be used to prove that limits exists (at least not in the way that I have presented it here), since it doesn’t cover any possible path of approach.
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