0.1 Limits and Continuity
Before we get into limits and continuity of multivariable functions, let’s firsts review what limits are for single variable functions. Please note that the contents here will not be as “formal” or rigorous as taught in Real Analysis courses, but maybe I could include them in the future! First, recall the following definition of limit for a single variable function.
Definition 0.1.1
Given a function \(f(x)\), the limit of \(f(x)\) as \(x\) approaches \(a\) denoted as \(\lim _{x \to a} f(x) = L\) is defined as the following:
Given any \(\epsilon > 0\), there exists \(\delta > 0\) such that if \(0 < | x - a | < \delta \), then \(| f(x) - L | < \epsilon \).
An intuitive way of understanding this epsilon-delta definition of limit would be finding a value that the bound of \(f(x)\) gets arbitrarily close to as the distance between the bound for \(x\) gets arbitrarily close to zero. Now for \(2\)-space, you are probably familiar with a graph involving a shaded rectangle. For \(3\)-space, it is similar, be we can think of it as a circle. Our bound is no longer the width of the rectangle, but the radius of the circle surrounding the point. Consider the following definition for functions in \(3\)-space, or functions with two variables.
Definition 0.1.2
Given a function \(f(x, y)\), the limit of \(f(x, y)\) as \((x, y)\) approaches \((x_0, y_0)\) denoted as \(\lim _{(x, y) \to (x_0, y_0)} f(x, y) = L\) is defined as the following:
Given any \(\epsilon > 0\), there exists \(\delta > 0\) such that for all \((x, y) \in \dom (f)\), if \(0 < \sqrt {(x - x_0)^2 + (y - y_0)^2} < \delta \), then \(| f(x, y) - L | < \epsilon \).
Considering that \(|x - a|\) is the distance between \(x\) and \(a\), I hope the expression \(\sqrt {(x - x_0)^2 + (y - y_0)^2}\) is a natural extension because it represents the distance between \((x, y)\) and \((x_0, y_0)\). Let’s take at an example.
Exercise 0.1.3
Show that \(\lim _{(x, y) \to (0, 0)} \frac {x + y}{x^2 + 1} = 0\).
Solution.
Given \(\epsilon > 0\), choose \(\delta = \frac {\epsilon }{\sqrt {2}}\). Then, the following equations hold by AM-GM inequality. \[ \left | \frac {x + y}{x^2 + 1} - 0 \right | \leq | x + y | \leq \sqrt {2x^2 + 2y^2} < \sqrt {2} \delta = \epsilon \] Therefore, \(0 < \sqrt {(x - x_0)^2 + (y - y_0)^2} < \delta \), then \(| f(x, y) - 0 | < \epsilon \).
Similarly, we can replace the points in \(2\)-space to \(3\)-space to generalize for function with three variables. Another way of thinking about limits for multivariable functions are through different curves.
Definition 0.1.4
Given a function \(f(x, y)\) and a smooth curve \(C\) in \(2\)-space with \(x(t) = x\) and \(y(t) = y\) and \((x_0, y_0) = (x(t_0), y(t_0))\), then the following equation hold. \[ \lim _{(x, y) \underset {\to }{C} (x_0, y_0)} f(x, y) = \lim _{t \to t_0} f(x(t), y(t)) \]
This is now easier because \(\lim _{t \to t_0} f(x(t), y(t))\) is now a single variable limit. Here is an example continuing with our previous one.
Exercise 0.1.5
Compute \(\lim _{(x, y) \to (0, 0)} \frac {x + y}{x^2 + 1}\) along the path \(y = x\).
Solution.
Notice that \(y = x\) can be represented as \(x(t) = t\) and \(y(t) = t\). Therefore, the limit can be rewritten as \(\lim _{t \to 0} \frac {t + t}{t^2 + 1} = 0\). Therefore, the limit along the path \(y = x\) is zero.
However an important thing to note is that we have infinitely many different ways to represent a limit along different curves because there are infinitely different curves. This can cause a problem. What if the limits return different values from one to another?
Finding limits along a curve does not represent the function’s limit and cannot be used to prove the limit.
This is important because it shows that we can use this method to disprove the existence of limit, but not find limit. Here is an example.
Exercise 0.1.6
Find, if any, the value of \(\lim _{(x, y) \to (0, 0)} \frac {3y}{x^2 + y}\).
Solution.
First, consider the limit along a curve \(y = kx^2\) for any \(k\). Notice that the limit can be represented as \(\lim _{t \to 0} \frac {3kx^2}{x^2 + kx^2} = \frac {3k}{1 + k}\). Because \(\frac {3k}{1 + k}\) depends on the value of \(k\) and is not constant, the limit does not exist.
One final thing before we wrap up our discussions about limit is that the properties of sum, different, product, and quotient applies the same for the generalized limits and limits along different curves just as the limit for single variable function!