Limits of rational functions

Define the function f : ℝ^{2} → ℝ by

f\binom{x}{y} = \frac{x^{2}y}{x^{4} + y^{2}} .

We want to consider the limit of this function at \binom{0}{0}: does it exist? What is it, if it does exist? Let’s consider first approaching the origin along a line. If the line is x = 0, then the function is constantly zero along this line, and so it has a limit along this line. Any other line has the equation y = mx, so we make this substitution. We get:

lim_{\binom{x}{y}→\binom{0}{0}} \frac{x^{2}(mx)}{x^{4} + (mx)^{2}} = lim_{\binom{x}{y}→\binom{0}{0}} \frac{mx^{3}}{x^{4} + m^{2}x^{2}} = lim_{\binom{x}{y}→\binom{0}{0}} \frac{mx}{x^{2} + m^{2}} = 0.

The last equality follows immediately if m = 0, and if m ≠ 0, then the limitand is continuous in x at the origin, so we can just substitute.

But what happens along a different path, say y = mx^{2}? Then the limit becomes

lim_{\binom{x}{y}→\binom{0}{0}} \frac{x^{2}(mx^{2})}{x^{4} + (mx^{2})^{2}} = lim_{\binom{x}{y}→\binom{0}{0}} \frac{mx^{4}}{x^{4} + m^{2}x^{4}} = lim_{\binom{x}{y}→\binom{0}{0}} \frac{m}{1 + m^{2}} ,

which depends on m. For example, if m = 0, then it’s zero, while if m = 1, it’s 1/2.

So, even though f has a limit along every line approaching the origin, and the limits are all the same along these lines, f itself does not have a limit at the origin. A limit of a function of 2 (or more) variables must be the same regardless of the method of approach.

By similar means, you should be able to show the following proposition: Proposition. Define f : ℝ^{2} → ℝ by

f\binom{x}{y} = \frac{x^{d}y}{x^{2d} + y^{2}} .

Then the limit of f at \binom{0}{0} along any path y = p(x), where p is a polynomial of degree less than d, is zero. However, the limit of f along paths of the form y = mx^{d} varies with m. Hence, the limit of f at \binom{0}{0} does not exist.

There is a kind of limit that occurs often enough that it’s useful to know immediately that the limit is zero. As you saw in recitation this week, I struggled to come up with a viable statement of this result. This is one of those times you should understand the technique of the proof more than knowing exactly what the theorem is, and I expect to see you use the kind of inequalities that appear in the proof. Proposition. Suppose p : ℝ^{n} → ℝ is a polynomial in n variables, where every term has degree greater than d. Then

lim_{|x|→0} \frac{|p(x)|}{|x|^{d}} = 0.

Proof. The basic idea, and one you should internalize and recognize immediately, is that the length of any component of a vector is less than or equal to the length of the whole vector. That is,

|x_{i}| ≤ \sqrt{x_{1}^{2} + x_{2}^{2} + ⋯ + x_{n}^{2}} for all 1 ≤ i ≤ n.

By the triangle inequality, the absolute value of a polynomial is less than or equal to the sum of the absolute values of its terms. Therefore we can just show that the proposition is true for a monomial ax_{1}^{i_{1}} … x_{n}^{i_{n}}, and it generalizes immediately to any polynomial satisfying the given conditions. Our assumption is d′ = i_{1}+⋯+i_{n} > d. Then, by the above observation,

|ax_{1}^{i_{1}} … x_{n}^{i_{n}}| ≤ |a| (\sqrt{x_{1}^{2} + ⋯ + x_{n}^{2}})^{d′} = |a||x|^{d′}.

Thus the limitand is bounded above at all points by

\frac{|a||x|^{d′}}{|x|^{d}} = |a||x|^{d′−d}.

Because we have assumed d′ > d, this expression certainly approaches zero as |x| → 0. Therefore the original limit is zero. □