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Math Help - How to calculate this limit?

  1. #1
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    How to calculate this limit?

    I cant seem to figure out what to do to solve this problem. Can anyone please help?

    lim x--> infinity ((2e^-x - 6e^x) / 2e^x + e^-x)

    The answer is -3 but how?

    also....how would I do this one? (very similar prob)

    lim x-->0+ ((2 - e^-x) / e^1/x )
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  2. #2
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    Is this \displaystyle \lim_{x \to \infty}\left(\frac{2e^{-x} - 6e^x}{2e^x} + e^{-x}\right) or \displaystyle \lim_{x \to \infty}\frac{2e^{-x} - 6e^x}{2e^x + e^{-x}}?
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  3. #3
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    sorry for being unclear, it is the latter though
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    \displaystyle \frac{2e^{-x} - 6e^x}{2e^x + e^{-x}} = \frac{\frac{2}{e^x} - 6e^x}{2e^x + \frac{1}{e^x}}

    \displaystyle = \frac{\frac{2 - 6e^{2x}}{e^x}}{\frac{2e^{2x} + 1}{e^x}}

    \displaystyle = \frac{2-6e^{2x}}{2e^{2x} + 1}.


    Therefore \displaystyle \lim_{x \to \infty}\frac{2e^{-x} - 6e^x}{2e^x + e^{-x}} = \lim_{x \to \infty}\frac{2 - 6e^{2x}}{2e^{2x} + 1}.

    Since this goes to \displaystyle \frac{\infty}{\infty}, you can use L'Hospital's Rule...
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  5. #5
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    Hello, colerelm1!

    \displaystyle \lim_{x\to\infty} \frac{2e^{-x} - 6e^x}{2e^x + e^{-x}}

    The function is: . \displaystyle \frac{2e^{-x} - 6e^x}{2e^x + e^{-x}}

    Divide numerator and denominator by e^x\!:\;\;\dfrac{2e^{-2x} - 6}{2 + e^{-2x}}


    \displaystyle\text{Therefore: }\;\lim_{x\to\infty}\left(\frac{\frac{2}{e^{2x}} - 6}{2 + \frac{1}{e^{2x}}}\right) \;=\; \frac{0-6}{2+0} \;=\;-3

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