# Thread: Vertical asymptote in polar?

1. ## Vertical asymptote in polar?

Show that the polar curve r = 4 + 2sec(theta) has the line x = 2 as a vertical asymptote by showing that lim (r -> inf) x = 2

I don't understand what its asking for me to do, and how I should do it.

I have, however, shown that r = r + 4 without using limits, but it specifically asks for a limit.

2. Originally Posted by LeoBloom.
Show that the polar curve r = 4 + 2sec(theta) has the line x = 2 as a vertical asymptote by showing that lim (r -> inf) x = 2

I don't understand what its asking for me to do, and how I should do it.

I have, however, shown that r = r + 4 without using limits, but it specifically asks for a limit.
Convert it to rectangular co-ordinates and evaluate the limit?

3. I don't understand. Perhaps, I have forgotten how to do limits.

I would really appreciate a step by step answer, even though it seems that I just want the answer, I learn best from seeing an example. (Its not in my text book either)

4. Originally Posted by LeoBloom.
I don't understand. Perhaps, I have forgotten how to do limits.

I would really appreciate a step by step answer, even though it seems that I just want the answer, I learn best from seeing an example. (Its not in my text book either)
Did you convert it to rectangular co-ordinates first? IF so show me and I will help with the limit

5. Does this look right?

1 = 4/(sqrt(x^2 + y^2)) + 2/x

6. Originally Posted by LeoBloom.
Does this look right?

1 = 4/(sqrt(x^2 + y^2)) + 2/x
Yeah it does...I am pretty sure...now $1=\frac{4}{\sqrt{x^2+y^2}}+\frac{2}{x}$
How would you solve that for a limit?

7. Well, I'd like you to help me understand limits again. It's been a while since we solved for limits in class. So, does lim (r to inf) x = 2 mean that r approaches inf when x is = to 2? And if that is the case, how do I "solve" this because I am unsure how to show it. I'd think that we can plug in x = 2 but what would that do?

8. Originally Posted by LeoBloom.
So, does lim (r to inf) x = 2 mean that r approaches inf when x is = to 2?
No, it means that $x$ approaches $2$ as $r$ approaches $+\infty$. To evaluate this limit you need to have an expression of $x$ which only depends on $r$, acknowledged that $x=r\cos \theta$ and $r=4+\frac{2}{\cos \theta}$.

9. OMG, thank you guys! I think I got it now. Basically, 2/(r - 4) = cos(Theta). x = r(2/(r - 4)). Lim r -> inf of 2r/(r - 4). Using L'Hospital, you get 2/1 = 2. Did I do it right?!

10. Originally Posted by LeoBloom.
OMG, thank you guys! I think I got it now. Basically, 2/(r - 4) = cos(Theta). x = r(2/(r - 4)). Lim r -> inf of 2r/(r - 4). Using L'Hospital, you get 2/1 = 2. Did I do it right?!
good job

11. As much as I hate asking another similar question, this one seems harder to solve. r = sinTtanT show that it has the line x = 1 as a vertical asymptote. I assume its done the same way but I dont see how to change r solely to x as flyign squirrel described

12. The idea is the same, except that me use $x(T)$ and $y(T)$ instead : $\left\{ \begin{array}{l}x(T)=r\cos T=\sin^2T\\y(T)=r\sin T=\frac{\sin^3T}{\cos T}\end{array}\right.$

As you are told that the asymptote is the line $x=1$, you can find for which $T_0$ it will be reached solving $\lim_{T\to T_0}x(T)=1$. (here, the limit is useless because $x(T_0)$ exists but it will not always be true) Then, check that $\lim_{T\to T_0}y(T)=\pm \infty$ and conclude.