# A tricky limit from the Math 32A final

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.