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Math Help - Find the horizontal and vertical asymptotes of the curve.

  1. #1
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    Find the horizontal and vertical asymptotes of the curve.

    Find the horizontal and vertical asymptotes of the curve. (Round your answers to two decimal places.)

    I already solved for the y's
    y=0
    y=5

    But I'm stuck on x?
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  2. #2
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    Hello, rodjav305!

    Find the horizontal and vertical asymptotes of the curve: . y \:=\:\frac{5e^x}{e^x-5}
    Round answers to two decimal places.

    For horizontal asymptotes, find the limit of y as x\to\infty.

    We have: . y \:=\:\frac{5e^x}{e^x-5}
    Divide top and bottom by e^x\!:\;\;\frac{\dfrac{5e^x}{e^x}}{\dfrac{e^x}{e^x  } - \dfrac{5}{e^x}} \;=\;\frac{5}{1 - \dfrac{5}{e^x}}

    Then: . \lim_{x\to\infty}y \;=\;\lim_{x\to\infty}\left[\frac{5}{1-\frac{5}{e^x}}\right] \;=\;\frac{5}{1-0} \;=\;5


    The horizontal asymptote is: . \boxed{y \:=\:5.00}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    For vertical asymptotes, we want: . \text{denominator} \,=\,0

    So we have: . e^x-5 \:=\:0 \quad\Rightarrow\quad e^x \:=\:5 \quad\Rightarrow\quad x \:=\:\ln 5

    The vertical asymptote is: . \boxed{x \:=\:1.61}

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  3. #3
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    Dear Soroban

    I am new to calculus and thought that was a usefull concept to "for vertical asymptotes" set the denominator to zero..
    My question is how come it works out? Can I prove that somehow? Is it allways valid?
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  4. #4
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    Quote Originally Posted by Henryt999 View Post
    I am new to calculus and thought that was a usefull concept to "for vertical asymptotes" set the denominator to zero..
    My question is how come it works out? Can I prove that somehow? Is it allways valid?
    Go and look up the definition of a vertical asymptote ....
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  5. #5
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    Smile

    Quote Originally Posted by Henryt999 View Post
    I am new to calculus and thought that was a usefull concept to "for vertical asymptotes" set the denominator to zero..
    My question is how come it works out? Can I prove that somehow? Is it allways valid?
    take the following function f:x\mapsto \frac{1}{x} it's domain is \mathbb{R}\setminus \left \{ 0 \right \}.and that's means x=0 is a vertical asymptote.

    • You can see that at x=0, f is discontinuous.
    • When we get closer to 0 from left and right,the curve get closer to the line x=0 and while doing that f takes values that go to infinity.


    1. \lim_{x\to 0^{+}}\left (\frac{1}{x}  \right )=+\infty
    2.  \lim_{x\to 0^{-}}\left (\frac{1}{x}  \right )=-\infty
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