Asymptote of a curve in polar coordinates

[Faiq]

Homework Statement

The curve $C$ has polar equation $r\theta =1$ for $0<\theta<2\pi$

Use the fact that $\lim_{\theta \rightarrow 0}\frac{sin \theta }{\theta }=1$ to show the line $y=1$ is an asymptote to $C$.

The Attempt at a Solution

**Attempt**

$$\ r\theta =1$$
$$\ y\frac{\theta}{\sin \theta} =1$$
$$\ y=\frac{\sin\theta}{ \theta} $$
$$\ \lim_{\theta \rightarrow 0}y= \lim_{\theta \rightarrow 0}\frac{sin \theta }{\theta }=1 $$

I understood the concept behind how this asymptote is calculated, but I am not very fluent in mathematics to convert the above information into a comprehensive proof.

Moreover, there is another statement that states that I have to make use of the information $\lim_{\theta \rightarrow 0}x=+\infty$.
Although I get the basic idea that for the line $y=1$ to be an asymptote, it should continue until x goes to infinity, I am not sure how is this $\lim_{\theta \rightarrow 0}x=+\infty$ derived (why not negative infinity ?)* and not even sure how to add it in the proof. And lastly, why is proving this information necessary?

Can someone sort my confusions out?

A little context for my confusion*

If I approach $\theta$ from negative side I get negative infinity, if from positive side I get positive infinity since $\lim_{\theta \rightarrow 0^\pm}x= \lim_{\theta \rightarrow 0^\pm}\frac{\cos \theta }{\theta } =\pm \infty$. So why are we just using positive infinity over here? Is it because of the positive range? If yes, what if the provided range was $-2\pi<\theta<2\pi$?

 

[Buzz Bloom]

**Attempt**

Hi Faiq:

I think there is a step missing in your proof. What is the relationship among: y, r, and θ?

Hope this helps.

Regards,
Buzz

 

[Mark44]

Homework Statement

The curve $C$ has polar equation $r\theta =1$ for $0<\theta<2\pi$

Use the fact that $\lim_{\theta \rightarrow 0}\frac{sin \theta }{\theta }=1$ to show the line $y=1$ is an asymptote to $C$.

The Attempt at a Solution

**Attempt**

$$\ r\theta =1$$
$$\ y\frac{\theta}{\sin \theta} =1$$
$$\ y=\frac{\sin\theta}{ \theta} $$
$$\ \lim_{\theta \rightarrow 0}y= \lim_{\theta \rightarrow 0}\frac{sin \theta }{\theta }=1 $$

I understood the concept behind how this asymptote is calculated, but I am not very fluent in mathematics to convert the above information into a comprehensive proof.

Your work above looks OK to me to show that $\lim_{\theta \to 0} y = 1$

Moreover, there is another statement that states that I have to make use of the information $\lim_{\theta \rightarrow 0}x=+\infty$.
Although I get the basic idea that for the line $y=1$ to be an asymptote, it should continue until x goes to infinity, I am not sure how is this $\lim_{\theta \rightarrow 0}x=+\infty$ derived (why not negative infinity ?)* and not even sure how to add it in the proof. And lastly, why is proving this information necessary?

Can someone sort my confusions out?

You are given that $0 < \theta < 2\pi$, which has an effect on what you get for $\lim_{\theta \to 0} x$. Use a strategy similar to the one you used to find a horizontal asymptote.

A little context for my confusion*

If I approach $\theta$ from negative side I get negative infinity, if from positive side I get positive infinity since $\lim_{\theta \rightarrow 0^\pm}x= \lim_{\theta \rightarrow 0^\pm}\frac{\cos \theta }{\theta } =\pm \infty$. So why are we just using positive infinity over here? Is it because of the positive range? If yes, what if the provided range was $-2\pi<\theta<2\pi$?

 

[Faiq]

You are given that 0<θ<2π0<θ<2π0 < \theta < 2\pi, which has an effect on what you get for limθ→0xlimθ→0x\lim_{\theta \to 0} x. Use a strategy similar to the one you used to find a horizontal asymptote.

Is this correct ?
$$\ x= \frac{\cos θ}{θ} $$
$$\ \lim_{θ→0}x= \lim_{θ→0} \frac{\cos θ}{θ}= +\infty $$
Can you also answer me what was I supposed to do if provided range was $−2π<θ<2π$ instead of $0<θ<2π$ ?

 

[Mark44]

Is this correct ?
$$\ x= \frac{\cos θ}{θ} $$
$$\ \lim_{θ→0}x= \lim_{θ→0} \frac{\cos θ}{θ}= +\infty $$

The last part should be $\lim_{θ→0^+} \frac{\cos θ}{θ}= +\infty$, since $0 < \theta < 2\pi$

Can you also answer me what was I supposed to do if provided range was $−2π<θ<2π$ instead of $0<θ<2π$ ?

The the limit above wouldn't exist, because the left-hand and right-hand limits would be different.

 

[Faiq]

The last part should be $\lim_{θ→0^+} \frac{\cos θ}{θ}= +\infty$, since $0 < \theta < 2\pi$
The the limit above wouldn't exist, because the left-hand and right-hand limits would be different.

Is it not possible to divide the range $-2\pi <\theta< 2\pi$ into $-2\pi <\theta< 0$ and $0 <\theta< 2\pi$?
Because it is evident that no matter what the range is the asymptote will always be present.

 

[Mark44]

Is it not possible to divide the range $-2\pi <\theta< 2\pi$ into $-2\pi <\theta< 0$ and $0 <\theta< 2\pi$?

On the given interval, $0 <\theta< 2\pi$, $\lim_{θ→0^+} \frac{\cos θ}{θ}= +\infty$.
On the other interval, $-2\pi <\theta< 0$, $\lim_{θ→0^-} \frac{\cos θ}{θ}= -\infty$.

Because it is evident that no matter what the range is the asymptote will always be present.

No, because there are two asymptotes.