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Math Help - horizontal asymptotes

  1. #1
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    horizontal asymptotes

    Hi again

    f(x) = (√(x^2+1)/(x-2))

    I think the vertical asymptote is 2 but what is the horizontal asymptote?

    How do you figure this out?

    Thanks in advance
    this calculus beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi again

    f(x) = (√(x^2+1)/(x-2))

    I think the vertical asymptote is 2 but what is the horizontal asymptote?

    How do you figure this out?

    Thanks in advance
    this calculus beginner
    To find the horizontal asymptotes you need to compute:
    \lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x-2}\text{ and }\lim_{x\to -\infty}\frac{\sqrt{x^2+1}}{x-2}.

    Can you compute those limits?
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  3. #3
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    I am finding it hard to deal with the square root signs so if you can walk me through it I would appreciate it.

    Thanks
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  4. #4
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    Quote Originally Posted by calcbeg View Post
    I am finding it hard to deal with the square root signs so if you can walk me through it I would appreciate it.

    Thanks
    \frac{\sqrt{x^2+1}}{x-2} = \frac{\sqrt{x^2+1}}{x-2} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}} = \frac{\sqrt{1 + \tfrac{1}{x^2}}}{1 - \tfrac{2}{x}} for x>0.
    What happens when x\to \infty?

    If x<0 then we need to be careful we cannot multiply both the numerator and denominator with \tfrac{1}{x} because \tfrac{1}{x} and you cannot put a negative number under a square root. Rather you need to multiply the denominator by \tfrac{1}{-x}. Can you do the second limit?
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  5. #5
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    The examples I have found seem to show getting rid of the x's on the top of the formula but you haven't done that with the last one. I am expecting to come up with a number and I don't know what that is on the first one let alone the second one. Do I substitute 0 for x at some point?

    Thanks
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  6. #6
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    Quote Originally Posted by calcbeg View Post
    The examples I have found seem to show getting rid of the x's on the top of the formula but you haven't done that with the last one. I am expecting to come up with a number and I don't know what that is on the first one let alone the second one. Do I substitute 0 for x at some point?

    Thanks
    The meaning of x\to \infty means x gets almost as large as my ego.
    Thus, if x is getting large it would mean \tfrac{1}{x} goes to zero, as well as \tfrac{1}{x^2}\to 0.

    Now, \lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x-2} = \lim_{x\to \infty}\frac{\sqrt{x^2+1}}{x-2} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}} = \lim_{x\to \infty} \frac{\sqrt{1 + \tfrac{1}{x^2}}}{1 - \tfrac{2}{x}} = \frac{\sqrt{1+0}}{1-0} = 1
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