horizontal asymptote

• Aug 16th 2008, 09:52 PM
11rdc11
horizontal asymptote
Ok this may be a dumb question but when you take the limit of a function to infinity you find out the horizontal asymptote correct? So my question is does sin(x)/x have a horizontal asymptote at 0? I got 0 being the horizontal asymptote using hopital rule. I also used the squeeze theorem and it gave the limit of sin(x) to infinity equals 0. What am I doing wrong becuase when I graph the function there is no asymptote? thanks
• Aug 16th 2008, 10:21 PM
o_O
Note by the definition you are using, it says nothing about not crossing the horizontal asymtote. In fact, it can oscillate about it infinitely many times such as your function $y = \frac{\sin x}{x}$ or just once such as $y = \frac{x}{1+x^2}$ or $y = \frac{3x^2 - x - 2}{5x^2 + 4x + 1}$.
• Aug 16th 2008, 10:57 PM
11rdc11
Thanks but im still a little confused I thought a asymptote was a break in the graph that a function may approach but never touches? Such as 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0. The left hand limit from 0 approaches neg infinity and the right hand limit form 0 approaches pos infinity. Also if I take the the limit of 1/x to infinity it approaches 0. I thought it could approach but not touch the value of the asymptote. So a function is allowed to have a point on the asymptote as in sin(x)/x when x=pi and if so why? Thanks.

P.S. Does convergance have anything to do with this?
• Aug 16th 2008, 11:12 PM
o_O
This goes back to the definition of a horizontal asymptote: The line y = a is a horizontal asymptote if $\lim_{x \to \infty} f(x) = a$ or $\lim_{x \to -\infty} f(x) = a$. Again, nothing is said about not being able to cross it. It is just a matter of what the function tends to as x approaches infinity.

Take a look at the examples that I gave you. Here are visual representations of the graphs to give you an idea of what I'm talking about: Limits at Infinity (Scroll down a bit)
• Aug 16th 2008, 11:28 PM
11rdc11
Thank you now it makes sense to me. I was taught that an aysmptote meant that a function could not touch that value but I now I understand it at the end points when taken to infinity that I need to pay attention to. Once again thank you and now that also solves question I had about curve sketching. Now back to trying to understand infinite series(Happy)
• Aug 16th 2008, 11:32 PM
mr fantastic
Quote:

Originally Posted by 11rdc11
Ok this may be a dumb question but when you take the limit of a function to infinity you find out the horizontal asymptote correct? So my question is does sin(x)/x have a horizontal asymptote at 0? I got 0 being the horizontal asymptote using hopital rule. I also used the squeeze theorem and it gave the limit of sin(x) to infinity equals 0. What am I doing wrong becuase when I graph the function there is no asymptote? thanks

You're attempting to draw a graph of the sinc function. See here: Sinc function - Wikipedia, the free encyclopedia.

By the way, l'Hopital's rule does not work when trying to find the limit as x --> +oo of sin(x)/x since the limit as x --> oo of sin(x) does not exist. sin(x)/x --> 0 as x --> oo because sin(x) is bounded .....
• Aug 16th 2008, 11:59 PM
11rdc11
Thank you Mr. F for pointing out why I can't use hopital rule. just to make sure I understand this right sin(x) to the limit of infinity does not exist because sin(x) never converges to anything. It could range from -1 to 1? That why to find the limit of sin(x)/x I would use the squeeze theorem correct? -1/x < or = or sin(x)/x < or = to 1/n. So essentially the squeeze theorem is like the monotonic sequence theorem just with an upper and lower bound? Thanks
• Aug 17th 2008, 02:04 AM
wingless
Quote:

Originally Posted by 11rdc11
Thank you Mr. F for pointing out why I can't use hopital rule. just to make sure I understand this right sin(x) to the limit of infinity does not exist because sin(x) never converges to anything. It could range from -1 to 1? That why to find the limit of sin(x)/x I would use the squeeze theorem correct? -1/x < or = or sin(x)/x < or = to 1/n. So essentially the squeeze theorem is like the monotonic sequence theorem just with an upper and lower bound? Thanks

Yes, It doesn't converge, it oscillates between -1 and 1.

The squeeze theorem says if there are functions $f(x)\leq g(x)\leq h(x)$, and if $\lim_{x\to a} f(x) = \lim_{x\to a} h(x)$, then $\lim_{x\to a} g(x) = \lim_{x\to a} f(x) = \lim_{x\to a} h(x)$.

Let $f(x) = -\frac{1}{x}$ and $h(x) = \frac{1}{x}$ then use the squeeze theorem as x goes to infinity.

http://img294.imageshack.us/img294/9644/gphhac0.png