Showing that $y = {\phi \over \phi -\pi} \cdot \sin\phi$ tends to $-\pi$ as $\phi \to \pi$ without using limits

Question

I'm trying to manually plot a parametric function:

$$ r=\frac{\phi}{\phi - \pi} $$

I've translated the above to Сartesian coordinate system and found zeros for $x$ and $y$ axes.

$$ \cases{ x = r\cos\phi = \frac{\phi}{\phi - \pi}\cdot \cos\phi\\ y = r\sin\phi = \frac{\phi}{\phi - \pi}\cdot \sin\phi } $$

Then i wanted to find the value of $x$ and $y$ as $\phi \to \pi$. It's clear that $x$ tends to $\pm \infty$ depending from which side $\phi$ approaches $\pi$ since $\cos\pi = -1$. But it's not clear how to deal with the value of $y$ since it's $y =\pm \infty\cdot0$.

I've found the limit using L'opitals rule, but the problem is taken from a precalculus section of the book i'm solving. I am not supposed to use limits or derivatives.

I've already spent a couple of hours pondering on that, but that doesn't seem very obvious to me.

So now I'm wondering how could I find that limit in a precalculus manner?

Below is the graph in case someone is curious what it's really about:

Answer 1

We can make use of the well-known fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and how it is proven. The main step in the prove is to note that $\cos x \le \frac{\sin x}{x} \le 1$. We will use this idea.

Now note that $\sin( \phi - \pi) = - \sin \phi $ and so we have:

$$ - \phi \le -\frac{\phi}{\phi-\pi} \sin(\phi -\pi) \le - \phi \cos(\phi - \pi)$$

Now you can see that the middle is bounded by $-\pi$ from both sides as we get closer and closer to $\pi$. Hence we can conclude that $y=-\pi$, as $\phi \to \pi$

Answer 2

The problem you describe is the very the definition of a limit.

$y(\pi)$ is not defined, but $y$ is defined in the neighborhood of $\pi$ and $y(\phi)$ is close to $\pi$ when $\phi$ is close to $\pi.$ That is a limit!

As this is pre-calc, and you don't have a formal definition of a limit, you are going to have to hand-waive to some degree. But, you don't need L'Hopitals rule.

You can evaluate $\frac {sin (\phi)}{\pi - \phi}$ via the squeeze theorem.

Probably easier to start with $\frac {\sin x}{x}$

$|\sin x| \le |x| \le |\tan x|$

And $|\sin (\pi - \phi)| \le |\pi - \phi| \le |\tan (\pi - \phi)|$

Geometrically:

The area of the large triangle is $\frac 12 \tan (\pi - \phi)$ The area of the section of the circle is $\frac 12 (\pi - \phi)$ And the area of the smaller traingle is $\frac 12 \sin (\pi - \phi)$

And with a little algebra:

$\cos (\pi - \phi)| \le \frac {\sin(\pi - \phi)}{\pi - \phi} \le 1$

and $\phi \cos (\pi - \phi)| \le \frac {\phi}{\pi - \phi}\sin (\pi-\phi) \le \phi$

Establishing an upper and lower bound, each close to $\pi$ as $\phi$ is close to $\pi$