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Math Help - limits toward neg. infinity

  1. #1
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    limits toward neg. infinity

    I understand that in order to solve, i would divide by the highest power of x. But what is he doing here that gives 7/7 for x > -infinity?




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  2. #2
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    Quote Originally Posted by Evan.Kimia View Post
    I understand that in order to solve, i would divide by the highest power of x. But what is he doing here that gives 7/7 for x > -infinity?




     \lim_{x \rightarrow - \infty  } \exp(x)=0
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  3. #3
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    Ah, makes perfect sense. Thank you.
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  4. #4
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    if i'm not mistaking your b question doesn't look right.
    \lim_{x \rightarrow  \infty  }  \frac{7-\exp x }{7+8\exp x } = \lim_{x \rightarrow  \infty  } \frac{\exp x( \frac{7 }{\exp x }-1)  }{\exp x( \frac{ 7}{\exp x }+8)  } = \lim_{x \rightarrow  \infty  } \frac{( \frac{7 }{\exp x }-1)  }{( \frac{ 7}{\exp x }+8)  } = -\frac{ 1}{8 }
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  5. #5
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    I'm sure for A he's using the fact that both -e^{x} and 8e^{x} towards negative infinity are both zero and that renders....

    \frac{7 - 0}{7 + 0}

    and B is correct.

    LOL. whoops, just saw the previous post.
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