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Math Help - horizontal asymptote

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    Super Member 11rdc11's Avatar
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    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
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    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}.
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    Super Member 11rdc11's Avatar
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    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?
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    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)
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    Super Member 11rdc11's Avatar
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    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
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    Quote Originally Posted by 11rdc11 View Post
    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 .....
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    Super Member 11rdc11's Avatar
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    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
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    Quote Originally Posted by 11rdc11 View Post
    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.

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