# Multivariable limit $\lim_{(x, y) \to (0, 0)} \frac{x \sin(x^2 y^2)}{x^2 + y^2}$ [duplicate]

I'm learning multivariable calculus, specifically multivariable limits and continuity, and need help to understand the solution to the following problem:

Let

$$f(x,y) = \begin{cases} \frac{x \sin(x^2 y^2)}{x^2 + y^2}, & (x,y) \neq (0, 0) \\ 0, & (x,y) = (0, 0). \end{cases}$$

Show that $f$ is continuous at $(0, 0)$.

So we need to show that

$$\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0)$$

that is

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

Solution: (from textbook)

We have that

$$\left| \frac{x \sin(x^2 y^2)}{x^2 + y^2} \right| = \frac{\left|x\right| \left|\sin(x^2 y^2)\right|}{x^2 + y^2} \leq \frac{\left|x\right|x^2 y^2}{x^2 + y^2} \leq \frac{\left|x\right|^3 y^2}{x^2 + y^2} \leq \left|x\right|^3.$$

Question:

I don't understand how the author found the upper bound for the sine function. Why does $\left|\sin(x^2y^2)\right| \leq x^2y^2$? When I first tried to solve this problem I used $\left|\sin(x^2y^2)\right| \leq 1$ without success. Can someone explein my error?

## marked as duplicate by user186170, user296602, Daniel W. Farlow, user99914, Alex ProvostAug 8 '16 at 5:18

• $| \sin x|\le |x|$ for all $x\in \mathbb R.$ – zhw. Aug 7 '16 at 16:13
For all $t$, $$|\sin(t)|\leq |t|.$$ Indeed, using mean value theorem, for all $t$, there is $|c_t|\leq |t|$ s.t. $$|\sin(t)|=|\sin(t)-\sin(0)|=\underbrace{|\cos(c_t)|}_{\leq 1}|t-0|\leq |t|$$
Basically, $$(t-\sin t)'=1-\cos t\ge0$$ then, since $0-\sin 0=0$, you get that $$t\ge \sin t, \quad t\ge0,$$ and by parity $$|\sin t|\le |t|,\quad t \in \mathbb{R}.$$
Can i point out another strategy? $$\lim_{(x, y) \to (0, 0)} \frac{x \sin(x^2 y^2)}{x^2 + y^2} = 0.$$ changing to polar coords: $$\lim_{\rho \to 0} \frac{\rho \cos t\, \sin(\rho^4 \cos^2t\,\sin^2t)}{\rho^2} = ... \textrm{taylor expansion} ... = 0$$ of course the method in the textbook is valid, proof for the sin inequality: $$|sin(t)|= \left| \int_0^t \cos(x) dx \right| \leq \left| \int_0^t \left| \cos(x) \right| dx \right| \leq \left|\int_0^t 1 dx \right| =|t|$$ so for all $t \in \mathscr R$: $$|sin(t)| \leq |t|$$