2
$\begingroup$

I read a similar question that was solved by using the fact that $|\sin(x)| \leq |x|$, but I tried a different approach, as I struggle with utilizing inequalities to solve limits of this kind. Please note that my textbook did not ask for the limit to be solved without using inequalities.

I attempted to prove this result by first switching to polar coordinates, then applying Hopital's rule as follows:

$$\lim_{(x,y) \to (0,0)} \frac{\sin^2 xy}{x^2 + y^2}=\lim_{\rho \to 0} \frac{\sin^2 (\rho^2\sin(\theta)\cos(\theta))}{\rho^2}$$

Now I use cosine duplication formula to rewrite $\sin^2(t)$

$$\lim_{\rho \to 0} \frac{\sin^2 (\rho^2\sin(\theta)\cos(\theta))}{\rho^2}=\lim_{\rho \to 0} \biggl(\frac{1}{2}\biggr)\frac{1-\cos(2\rho^2\sin(\theta)\cos(\theta)}{\rho^2}$$

and now I apply Hopital's rule

$$\lim_{\rho \to 0} \biggl(\frac{1}{2}\biggr)\frac{4\rho\sin(\theta)\cos(\theta)\sin(2\rho^2\sin(\theta)\cos(\theta))}{2\rho}=\lim_{\rho \to 0} \sin(\theta)\cos(\theta)\sin(2\rho^2\sin(\theta)\cos(\theta))$$

Now if I'm understanding this correctly, because the argument of the second sine function goes to zero, the result is proven.

Have I made any mistakes? Was there a faster or more intuitive approach to solving this problem without using inequalities?

$\endgroup$
5
  • 1
    $\begingroup$ Why do you want to use the duplication formula instead of performing directly the derivative on $\sin^2$? Anyway your result is correct, but let me say that every time you have the argument of a sin which tends to 0, the first approach is to write the first order approximation $\sin u = O(u)$. I don't know if this is what you call inequality, but once you're able to use it "naturally", it makes everything much easier. $\endgroup$
    – ECL
    Jun 2, 2020 at 10:24
  • $\begingroup$ Thanks! I didn't directly derive $\sin^2$ because I'm a bit rusty on my calc, and using duplication formula makes integrating $\sin^2$ much easier, so my brain went straight for that. Thank you for pointing it out though! Following up on your big-o observation: what you are saying is that I could have just written $\sin(\theta)\cos(\theta)\sin^2(\rho^2\sin(\theta)\cos(\theta)) \sim \rho^2\sin^2(\theta)\cos^2(\theta)$ ? $\endgroup$
    – Jack
    Jun 2, 2020 at 10:48
  • $\begingroup$ "without using inequalities" are you aware that the definition of limit in calculus necessitate the use of inequalities? So all the mathematical expressions written in your post heavily use inequalities. $\endgroup$
    – Surb
    Jun 2, 2020 at 12:52
  • $\begingroup$ @Surb I am, I wrote "without using inequalities" for lack of a better term. What I actually meant is "without defining a function that is greater or equal to the given function and prove that that function has the same limit we are trying to prove". In other words, without using a "squeeze theorem" approach $\endgroup$
    – Jack
    Jun 2, 2020 at 13:16
  • 1
    $\begingroup$ I meant $\sin(xy)\simeq xy$, i.e. $\sin^2(xy)\simeq x^2 y^2$, which then you can write in polar coordinates if you want. $\endgroup$
    – ECL
    Jun 2, 2020 at 13:24

1 Answer 1

Reset to default
3
$\begingroup$

Another way

First rewrite the expression: $$ \frac{\sin^2 xy}{x^2 + y^2} = \left( \frac{\sin xy}{xy} \right)^2 \frac{x^2y^2}{x^2+y^2} $$

Here the first factor tends to $1$ since $\frac{\sin t}{t} \to 1$ as $t \to 0$ and $xy \to 0$ as $(x,y) \to (0,0).$

For the second factor we use polar coordinates: $$ \frac{x^2y^2}{x^2+y^2} = \frac{\rho^4 \cos^2\theta \sin^2\theta}{\rho^2} = \rho^2 \cos^2\theta \sin^2\theta \to 0 $$ as $\rho \to 0$ which is the case when $(x,y) \to (0,0).$

Thus, $$ \frac{\sin^2 xy}{x^2 + y^2} \to 1^2 \cdot 0 = 0. $$

$\endgroup$
1
  • $\begingroup$ Thank you! I really like the way you solved it $\endgroup$
    – Jack
    Jun 2, 2020 at 13:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .