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Math Help - Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity

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    Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity

    Hey all,

    I been playing around with limits, when i solve for the following function:

    \lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1
    But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does \{e^{x}-1}=1+e^{x} ?. Thanks
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    Quote Originally Posted by Oiler View Post
    Hey all,

    I been playing around with limits, when i solve for the following function:

    \lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1
    But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does \{e^{x}-1}=1+e^{x} ?. Thanks
    ????

    If you graph that, you obtain a line y=1
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    Quote Originally Posted by dwsmith View Post
    ????

    If you graph that, you obtain a line y=1
    dwsmith, sorry just made some changes to the formula.
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    Quote Originally Posted by Oiler View Post
    dwsmith, sorry just made some changes to the formula.
    It is -1 when you go to negative infinity.
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  5. #5
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    Quote Originally Posted by Oiler View Post
    also how does \{e^{x}-1}=1+e^{x} ?. Thanks
    Let x = 0

    0 = 2 is that true?
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    ofcourse, how dim of me. Thanks dwsmith.
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    \displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}

    Applying L'Hopitals Rule
    \displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}

    Applying L'Hopitals Rule
    \displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1
    L'H˘pital's Rule is overkill for this. XD

    \lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1 since e^{-x}\rightarrow0 as x\rightarrow\infty.
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    Quote Originally Posted by Chris L T521 View Post
    L'H˘pital's Rule is overkill for this. XD

    \lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1 since e^{-x}\rightarrow0 as x\rightarrow\infty.
    Sometimes you just have to kill it.
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    \displaystyle \lim_{x \to \infty}\frac{e^x - 1}{e^x + 1} = \lim_{x \to \infty}1 - \frac{2}{e^x + 1}

    \displaystyle = 1 - 0

    \displaystyle = 1.
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    Quote Originally Posted by dwsmith View Post
    Sometimes you just have to kill it.
    But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.
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    Quote Originally Posted by Oiler View Post
    Hey all,

    I been playing around with limits, when i solve for the following function:

    \displaystyle\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1

    But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations?

    also how does \{e^{x}-1\}=\{1+e^{x}\} ?.

    Thanks
    For this fraction, the denominator is always 2 greater than the numerator.

    However, the "ratio" gets closer and closer to 1, as we increase x above 0.

    x\rightarrow\infty means x increases without bound above 0.

    Hence, as x "approaches infinity", you need infinitely many decimal places
    to express the value of the fraction as a value "other than 1".

    That's part of the concept of limits.

    viz-a-viz

    \displaystyle\frac{e^1-1}{e^1+1}=0.46211715726

    \displaystyle\frac{e^2-1}{e^2+1}=0.76159415596

    \displaystyle\frac{e^{10}-1}{e^{10}+1}=0.99990920426

    ....onward.

    If x<0 and decreases without bound below 0.

    x=-y

    x\rightarrow\ -\infty\Rightarrow\ y\rightarrow\infty

    \displaystyle\lim_{x \to -\infty}\frac{e^x-1}{e^x+1}=\lim_{y \to \infty}\frac{e^{-y}-1}{e^{-y}+1}=\lim_{y \to \infty}\frac{\frac{1}{e^y}-1}{\frac{1}{e^y}+1}

    \displaystyle\lim_{y \to \infty}\left[\frac{\frac{1}{e^y}}{\frac{1}{e^y}}\right]\;\frac{1-e^y}{1+e^y}\right]=-\lim_{y \to \infty}\frac{e^y-1}{e^y+1}=-1

    When you have the concept, you can apply the fast methods as you please later.
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    Quote Originally Posted by HallsofIvy View Post
    But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.
    To me, that was mighty simply. The constant disappear and the exponentials are their same derivatives.
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