The following limit appeared on yesterday’s Math 32A final:

\[

\lim_{(x, y) \to (0, 0)} \frac{\sin x^2 + \tan^2 y}{x^2 + \tan y^2}.

\]

The limit seemed to cause a lot of trouble. In fact, no one in the lecture of 200 students had a perfect solution! I thought I would write up a solution here to give some closure on the problem. I will use only tools that appear in a standard calculus course: the squeeze theorem and the fact that \(\lim_{x \to 0} \sin(x) / x = 1\).

The first thing to note is that evaluating the limit along the \(x\)-axis (taking \(y = 0\)) gives a limit of \(1\):

\[

\lim_{x \to 0} \frac{\sin x^2}{x^2} = \lim_{u \to 0} \frac{\sin u}{u} = 1.

\]

This tells us that *if the original limit exists*, it must be \(1\). However, this argument does not allow us to conclude that the limit does indeed exist. Choosing any other path through \((0, 0)\) will also give a limit of \(1\), so one should anticipate that the limit exists and is equal to \(1\). We remark that, in general, \(\lim_{(x, y) \to (a, b)} f(x, y) = L\) if and only if \(\lim_{(x, y) \to (a, b)} |f(x, y) – L| = 0\). This reformulation is convenient because \(0 \leq |f(x, y) – L|\), so applying the squeeze theorem is less fussy: we only need to bound \(|f(x, y) – L| \leq g(x, y)\) where \(\lim_{(x, y) \to (a, b)} g(x, y) = 0\). Thus, we wish to show that

\[

\lim_{(x, y) \to (0, 0)} \left| \frac{\sin x^2 + \tan^2 y}{x^2 + \tan y^2} – 1\right| = 0.

\]

To this end, we compute the following bound

\[

\begin{align*}

\left| \frac{\sin x^2 + \tan^2 y}{x^2 + \tan y^2} – 1\right| &= \left| \frac{\sin x^2 + \tan^2 y}{x^2 + \tan y^2} – \frac{x^2 + \tan y^2}{x^2 + \tan y^2}\right|\\

&= \left| \frac{\sin x^2 – x^2 + \tan^2 y – \tan y^2}{x^2 + \tan y^2}\right|\\

&\leq \left| \frac{\sin x^2 – x^2}{x^2 + \tan y^2}\right| + \left| \frac{\tan^2 y – \tan y^2}{x^2 + \tan y^2}\right|\\

&\leq \left| \frac{\sin x^2 – x^2}{x^2}\right| + \left| \frac{\tan^2 y – \tan y^2}{\tan y^2}\right|.

\end{align*}

\]

The first inequality holds by the triangle inequality (\(|a + b| \leq |a| + |b|\)), while the second inequality holds because \(\tan y^2 \geq 0\) (when \(y\) is close to \(0\)) and \(x^2 \geq 0\). Since we have

\[

0 \leq \left| \frac{\sin x^2 + \tan^2 y}{x^2 + \tan y^2} – 1\right| \leq \left| \frac{\sin x^2 – x^2}{x^2}\right| + \left| \frac{\tan^2 y – \tan y^2}{\tan y^2}\right|,

\]

it suffices to show that the limit of the expression on the right is \(0\) as \((x, y) \to (0, 0)\). The expression on the right is much easier to deal with than the original limit because it is the sum of two terms, each of which only involves a single variable. Thus, one can use familiar methods from single variable calculus to compute the limits. One can show that in fact we have

\[

\lim_{x \to 0} \left| \frac{\sin x^2 – x^2}{x^2}\right| = 0 \quad\text{and}\quad \lim_{y \to 0} \left| \frac{\tan^2 y – \tan y^2}{\tan y^2}\right| = 0.

\]

The first limit follows easily from the fact that \(\lim_{x \to 0} \sin(x) / x = 1\), while the second limit requires a little more ingenuity. We compute

\[

\begin{align*}

\lim_{y \to 0} \left| \frac{\tan^2 y – \tan y^2}{\tan y^2}\right| &= \left| \lim_{y \to 0} \frac{\tan^2 y}{\tan y^2} – \lim_{y \to 0} \frac{\tan y^2}{\tan y^2} \right|\\

&= \left| \lim_{y \to 0} \frac{\tan^2 y}{\tan y^2} – 1 \right|\\

&= \left| \lim_{y \to 0} \frac{\tan^2 y}{y^2} \frac{y^2}{\tan y^2} – 1 \right|\\

&= \left| \lim_{y \to 0} \left(\frac{\tan y}{y}\right)^2 \frac{y^2}{\tan y^2} – 1 \right|\\

&= | (1)^2 (1) – 1 |\\

&= 0.

\end{align*}

\]

The penultimate equality follows from the fact that \(\lim_{y \to 0} \tan(y) / y = 1\), which can be deduced by writing \(\tan y = (\sin y) / (\cos y)\) and using the familiar \(\sin(y) / y\) limit identity (or an application of L’Hopital’s rule). At any rate, we may now deduce from the squeeze theorem that

\[

\lim_{(x, y) \to (0, 0)} \frac{\sin x^2 + \tan^2 y}{x^2 + \tan y^2} = 1.

\]

Whew.