# Homework Help: Vertical Asymptote

1. Jan 29, 2006

### gator

Question: Find the Vertical Asymptote if any

f(x)=[tex}\frac{2x}{\sin2x}[\tex]

Im somewhat lost, I cross multi and get 1/2 although my friends tell me this just equals 1. Any ideas?

err....cant get the cool math symbol things to work.....

Last edited: Jan 29, 2006
2. Jan 29, 2006

### TD

Could you clarify the function? The LaTeX didn't work properly, check your tags.

3. Jan 29, 2006

### gator

im trying to say f(x)= 2x / sin2x

4. Jan 29, 2006

### TD

Use the correct tags (closing with /, not \) and click to see the source:

$$\frac{2x}{\sin2x}$$

Now, you can encounter a vertical asymptote if the denominator becomes zero for some value of x while the nominator is non-zero there.
If both are zero, you can check with a limit what the function becomes.

5. Jan 29, 2006

### gator

The only thing that would have the denom zero would be zero (sin2(o)) = 0
but that would cause the nom to become zero

6. Jan 29, 2006

### TD

Both become zero there indeed, in which case you can't tell (yet) if there will be a vertical asymptote. Have you covered limits?
You forgot one thing though, the sine-function becomes zero somewhere else (of course, it's period with period $2\pi$, but it has another zero within [0,$2\pi$]). Think about the trigonometric unit circle, when is the sine zero as well?

7. Jan 29, 2006

### gator

sin = zero @ pi (180* on unit circle)

yep, covered limits. But in our notes he covers finding the VA without using limits (weird i know) so im not sure

8. Jan 29, 2006

### TD

Indeed, at $x = \pi$ as well! And the nominator doesn't become zero there, so we have a real vertical asymptote here.
Of course, don't forget the sin(x) is periodic, so not only at $x = \pi$ but also at...? And the nominator only became zero at x = 0, but the denominator became zero there, so also at multiples of $2 \pi$ where the nominator doesn't become zero anymore!
Can you follow?

Perhaps you saw a standard limit you may use for the 0/0 case, which is the well-known:

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$

9. Jan 29, 2006

### gator

So my VA is x=pi ?

10. Jan 29, 2006

### TD

There's one yes, but there are more. Reread my last reply, I edited while you were posting.

11. Jan 29, 2006

### gator

sin2x is zero when x=1/2 pi ? since that would mean sin = pi which would be zero on the denom

12. Jan 29, 2006

### TD

Well, sin(x) is a periodic function with period $2\pi$. This means that if it becomes zero at x = 0, it also becomes zero when you add a multiple of $2\pi$. The same goes for the case $x = \pi$.

So we have that sin(x) becomes zero at: $x = 0 + 2k\pi \vee x = \pi + 2k\pi$ where k is an integer. Now, the nominator is non-zero in all those cases, except for x = 0 (but not for its multiples of $2\pi$). So if we leave out x = 0 in the x values I listed, then we have all the places where a vertical asymptote occurs. You see?

13. Jan 29, 2006

### gator

ok, i have you until the last sentsnce. if "k" is an interger, then the amount of VA's i could have is endless.

Im also a little lost on what i should write down on my assignment

by the way, thanks for all this help

14. Jan 29, 2006

### TD

Well, you're right! This function has an infinite number of vertical asymptotes!

15. Jan 29, 2006

### gator

lol yea!

now are you any good at continuity? I have a question

using the def. of continuity to determine the points at which the function is no continous and indicate whether the discontinuity is removable or not

f(x) = HUGE bracket (x+1)absolute value x-2 all over 2 sqrt x+1 x<0
1 x=0
-e^x / x+2 x>0

I have no notes on continuity, so if you could just lead me to a good site if you dont have time to help then that fine

16. Jan 29, 2006

### gator

and a quick question on limits

lim x right arrow 0 e^3x -1 / e^x -1

since anything to the root 0 is 1 you get 1-1/1-1 = 0

17. Jan 29, 2006

### TD

So you're looking for

$$\mathop {\lim }\limits_{x \to 0} \frac{{e^{3x} - 1}}{{e^x - 1}}$$

Have you seen L'Hopital's rule? Or can you factor the nominator knowing that $e^{3x} - 1 = \left( {e^x } \right)^3 - 1^3$?

Last edited: Jan 29, 2006
18. Jan 29, 2006

### gator

L'Hopital's rule?

19. Jan 29, 2006

### TD

Oh I see you made a mistake: (1-1)/(1-1) is 0/0 which isn't the same as 0...! You encounter an 'indeterminate form' here, namely 0/0 where you can apply L'Hopital's rule which states that in this case, you can take the nominator's and denominator's derivative (seperately, so f/g becomes f'/g'). If you don't know this rule, try factoring so you'll be able to cancel out $e^x-1$.

20. Jan 29, 2006

### gator

so like (e^x - 1)(e^x2 -e^x +1) so I now have e^x2 - e^x +1

Last edited: Jan 29, 2006