limits of 2 variables with trigonometric terms

Question

I'm trying to determine the following limit

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

I tried to use polar coordinates and then Taylor and got

$$ \lim_{r \to 0} \frac{2-2 \cdot \cos^2(\theta) \cdot \sin^2(\theta) \cdot \cos(r^2\cos^2(\theta)\sin^2(\theta))}{2+2 \cdot \cos^2(\theta) \cdot \sin^2(\theta) \cdot \cos(r^2\cos^2(\theta)\sin^2(\theta))}$$

Which does not say anything I guess? How can I solve this type of questions?

Answer 1

We have $ 0\le \sin t\le t,\ 0\le t\le {\pi \over 2}.$ Thus $${x^2+y^2-x^2y^2\over x^2+y^2+x^2y^2}\le h(x,y)\le 1\quad (*)$$ As $x^2y^2\le (x^2+y^2)^2$ we get $${1-(x^2+y^2)\over 1+(x^2+y^2) }\le h(x,y)\le 1$$ Thus $\displaystyle\lim_{(x,y)\to (0,0)}h(x,y)=1.$

Remark We can as well apply the polar coordinates to LHS of $(*).$ Then $${x^2+y^2-x^2y^2\over x^2+y^2+x^2y^2}={1-r^2\sin^2\theta\cos^2\theta\over 1+r^2\sin^2\theta\cos^2\theta}\to 1, \ r\to 0^+$$

Answer 2

Let $f(x,y) = x^2 - \sin(x^2 y^2) + y^2$ and $g(x,y) = x^2 + \sin(x^2 y^2) + y^2$, and $h(x,y) = f(x,y)/g(x,y)$. Right away, we can see that $h(\pm x, \pm y) = h(|x|,|y|)$, so the behavior of $h$ at $(0,0)$ can be investigated without loss of generality by assuming $x \ge 0$ and $y \ge 0$. We can also see that $h(y,x) = h(x,y)$, so we can further assume, for instance, $0 \le y \le x$.

Next, observe $h(x,0) = h(0,y) = 1$, so if the limit exists, it must be $1$. As you have done, let $x = r \cos \theta$, $y = r \sin \theta$, where $0 \le \theta \le \pi/4$; hence

$$h(x,y) = \frac{r^2 - \sin \left(r^4 (1-\cos 4\theta)/8\right)}{r^2 + \sin \left(r^4 (1 - \cos 4\theta)/8\right)} = -1 + \frac{2}{1 + \frac{\sin (\lambda r^4)}{r^2}}$$ where $\lambda = (1 - \cos 4\theta)/8$. Since $0 \le \theta \le \pi/4$, it follows that $0 \le \lambda\le \frac{1}{4}$, hence $$0 \le \frac{\sin (\lambda r^4)}{r^2} \le \frac{\sin (r^4/4)}{r^2}.$$ Now taking the limit as $r \to 0^+$ yields $$0 \le \lim_{r \to 0^+} \frac{\sin (\lambda r^4)}{r^2} \le 0,$$ hence $$\lim_{(x,y) \to 0} h(x,y) = -1 + \frac{2}{1} = 1.$$

Technically we didn't need to do the extra work of restricting $0 \le \theta \le \pi/4$, but in my mind, it makes the argument a little tidier at the end, and it is in general a good practice to exploit symmetry when possible. The key insight here is to observe that no matter if we allow $\theta$ to vary as a function of $r$, the value of $\lambda$ must remain bounded within a finite interval, which is what allows us to bound the corresponding limit.

Answer 3

We have that for $xy=0$ the limit is equal to $1$, for $xy\neq 0$ we have

$$ \frac{x^2-\sin(x^2y^2)+y^2}{x^2+\sin(x^2y^2)+y^2}=1-2\frac{\sin(x^2y^2)}{x^2y^2}\frac{x^2y^2}{x^2+\sin(x^2y^2)+y^2}\to 1$$

indeed $\frac{\sin(x^2y^2)}{x^2y^2}\to1$ and

$$0\le \left|\frac{x^2y^2}{x^2+\sin(x^2y^2)+y^2}\right|\le \frac{x^2y^2}{x^2+y^2}$$

with

$$\frac{x^2y^2}{x^2+y^2}\to 0$$