For a point $P(a,b)$ is a point lying on the curve satisfying $$2xy^2dx + 2x^2 y dy - \tan(x^2y^2) dx =0 $$

$\lim_{a\to -\infty}b = ? $

Options are:
a) $ 0$

b) $-1 $

c) $1$

d) does not exist.


If we observe carefully we get:

$d(x^2 y^2) = \tan(x^2 y^2) dx$

$\implies \ln(c\sin x^2y^2) = x$

$\implies c \sin (x^2 y^2) = e^x$

Now clearly as $x \to -\infty ~ , e^x \to 0$, so clearly $y \to 0$

But answer given is d. Please let me know my mistake.

  • $\begingroup$ I’ve never taken dif. eq., but my first instinct tells me that you’re going to have a constant of integration there somewhere, so how could you evaluate the limit to a specific number? $\endgroup$ – gen-z ready to perish Sep 30 '18 at 15:40
  • $\begingroup$ @ChaseRyanTaylor for example.what is the limit as $x\to\infty$ of the solution of $y'(x)=y(x)$? The solution will be $\infty$ no matter what the constant will be $\endgroup$ – ℋolo Sep 30 '18 at 18:56
  • $\begingroup$ @Holo $\infty$ is not a specific number in $\Bbb R$, it’s a direction $\endgroup$ – gen-z ready to perish Sep 30 '18 at 22:02
  • $\begingroup$ @ChaseRyanTaylor it is a limit, and we can extend $\Bbb R$ to include $\infty$. Also it is possible to create an example for any number there is $\endgroup$ – ℋolo Sep 30 '18 at 22:08

We have the following line:

$$c \sin (x^2 y^2) = e^x$$

The problem is that $\arcsin$ return only a single branch of $\sin$, not all of them:

Now, $c \sin (x^2 y^2) = e^{x}$ gives $$\sin (x^2 y^2) = \frac{e^{x}}c\implies x^2 y^2=\arcsin\left(\frac{e^{x}}c\right)\color{red}{+2k\pi}\\\implies y=\pm\frac1x\sqrt{\arcsin\left(\frac{e^{x}}c\right)\color{red}{+2k\pi}}$$So you have more then a single value for $y$ for any fixed $x$ and the answered are as big as we want so the limit won't get rid from them. So the limit is undefined

  • 2
    $\begingroup$ Still, for any fixed solution function (choice of $\pm$ and $c$ and $k$), the limit is $0$. But the "curve" in general is a collection of all such functions (with fixed $c$ only), so I think you are correct. $\endgroup$ – mr_e_man Sep 30 '18 at 18:40
  • 2
    $\begingroup$ Only non-negative $k$ are allowed, because of the $\sqrt\;$, and $\arcsin\in[-\pi/2,\pi/2]$. $\endgroup$ – mr_e_man Sep 30 '18 at 18:45
  • $\begingroup$ @mr_e_man you are correct that for every fixed choice, but we don't choose what $k$ is, the collection as a whole is $\{\pm\frac1x\sqrt{\arcsin\left(\frac{e^{x}}c\right)\color{red}{+2k\pi}}\mid k\in\Bbb N\}$, we don't choose one, you need that all of them to go together, or in the words of the definition of limit, you need that if you choose $\varepsilon>0$ there exists $N$ such that for all $x<N$ we will have that all of the branches will be less than $\varepsilon$(in absolute value), and this is impossible $\endgroup$ – ℋolo Sep 30 '18 at 18:49
  • $\begingroup$ Yes, that's what I said. $\endgroup$ – mr_e_man Sep 30 '18 at 18:50
  • $\begingroup$ @mr_e_man I see, I misunderstood, so your observation is on point $\endgroup$ – ℋolo Sep 30 '18 at 18:51

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