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Math Help - Easy yet tricky limit question (showing my work)

  1. #1
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    Easy yet tricky limit question (showing my work)

    I have to show my work on a problem: lim_{x \to -\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}

    Ok, so first thing: divide by fastest growing term in the denominator, which in this case as x approaches -\infty is actually: e^{-x}
    So: lim_{x \to -\infty}\frac{\frac{e^x}{e^{-x}}-\frac{e^{-x}}{e^{-x}}}{\frac{e^x}{e^{-x}}+\frac{e^{-x}}{e^{-x}}} = lim_{x \to -\infty}\frac{e^{2x}-1}{e^{2x}+1}

    This is where I start to get a little confused. How exactly do I mathematically express that as x approaches -\infty e^{2x} approaches zero?

    I can say: \frac{lim_{x \to -\infty}e^{2x}-1}{lim_{x \to -\infty}e^{2x}+1}, but then what? I have seen direct substitution with -\infty but isn't this bad form? You can't do arithmetic on infinity.

    So then does it just "jump" to: \frac{0-1}{0+1} = \frac{-1}{1} = -1?

    Or is there some merit to doing this next (after splitting the func. to a quotient of limits): \frac{e^{2*-\infty}-1}{e^{2*-\infty}+1} = \frac{e^{-\infty}-1}{e^{-\infty}+1}? I thought this had no meaning; if it does what is the next step?

    Thank you for any help, I know I got the answer, but it's the showing of the steps I am a little tripped up on. IF there is anything missing, can you also explain then how I can exactly express the other horizontal limit lim_{x \to \infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}?
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  2. #2
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    Re: Easy yet tricky limit question (showing my work)

    Quote Originally Posted by rortiz81 View Post
    I have to show my work on a problem: lim_{x \to -\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}

    Ok, so first thing: divide by fastest growing term in the denominator, which in this case as x approaches -\infty is actually: e^{-x}
    So: lim_{x \to -\infty}\frac{\frac{e^x}{e^{-x}}-\frac{e^{-x}}{e^{-x}}}{\frac{e^x}{e^{-x}}+\frac{e^{-x}}{e^{-x}}} = lim_{x \to -\infty}\frac{e^{2x}-1}{e^{2x}+1}

    This is where I start to get a little confused. How exactly do I mathematically express that as x approaches -\infty e^{2x} approaches zero?

    I can say: \frac{lim_{x \to -\infty}e^{2x}-1}{lim_{x \to -\infty}e^{2x}+1}, but then what? I have seen direct substitution with -\infty but isn't this bad form? You can't do arithmetic on infinity.
    No, you shouldn't just replace x with "infinity" but you can use the basis limit "properties":
    \lim_{x\to-\infty} \frac{e^{2x}- 1}{e^{2x}+ 1}=  \frac{\lim_{x\to -\infty} (e^{2x}- 1)}{\lim_{x\to -\infty} (e^{2x}+ 1)}=  \frac{\lim_{x\to -\infty}e^{2x}- \lim_{x\to -\infty} 1}{\lim_{x\to -\infty}e^{2x}+ \lim_{x\to -\infty}1}
    And now you can replace those limits with their values to get \frac{0- 1}{0+ 1}.

    So then does it just "jump" to: \frac{0-1}{0+1} = \frac{-1}{1} = -1?

    Or is there some merit to doing this next (after splitting the func. to a quotient of limits): \frac{e^{2*-\infty}-1}{e^{2*-\infty}+1} = \frac{e^{-\infty}-1}{e^{-\infty}+1}? I thought this had no meaning; if it does what is the next step?

    Thank you for any help, I know I got the answer, but it's the showing of the steps I am a little tripped up on. IF there is anything missing, can you also explain then how I can exactly express the other horizontal limit lim_{x \to \infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}?
    Last edited by HallsofIvy; July 2nd 2012 at 04:58 AM.
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  3. #3
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    Re: Easy yet tricky limit question (showing my work)

    Thank you, I skipped the sum/difference of limits law in both the numerator and denominator for brevity, but I will include that in my work.
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