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clara1223

  • one year ago

find the limit as x approaches infinity of (4x-7)/((5x^2)+3x+7)

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  1. clara1223
    • one year ago
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    \[\lim_{x \rightarrow \infty}\frac{ 4x-7 }{ 5x ^{2}+3x+7 }\]

  2. clara1223
    • one year ago
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    We haven't gone over limits as x approaches infinity in class before.

  3. amistre64
    • one year ago
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    divide it all by the highest power of x to simplify. anything left with an x in the denominator will zero out since 1/inf is a very very small amount of cake to get

  4. clara1223
    • one year ago
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    so multiply the denominator by 1/x^2?

  5. amistre64
    • one year ago
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    your basically searching for any horizontal asymptotes ... you covered them yet?

  6. amistre64
    • one year ago
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    top and bottom yes

  7. amistre64
    • one year ago
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    there are 'rules' which you can commit to memory if you have the gigabytes to play with :)

  8. clara1223
    • one year ago
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    so I'm left with \[\frac{ \frac{ 4 }{ x }-\frac{ 7 }{ x ^{2} } }{ 5+\frac{ 3 }{ x }+\frac{ 7 }{ x ^{2} } }\]

  9. amistre64
    • one year ago
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    now everything with an x under it, goes to something very very small, they approach zero. what are we left with?

  10. clara1223
    • one year ago
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    1/5?

  11. amistre64
    • one year ago
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    powers of x that is ... 4/x - 7/x^2 is not 1

  12. clara1223
    • one year ago
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    -1?

  13. amistre64
    • one year ago
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    0-0 = ??

  14. clara1223
    • one year ago
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    oh so the answer is 0?

  15. amistre64
    • one year ago
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    yep

  16. clara1223
    • one year ago
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    because 0/5=0

  17. amistre64
    • one year ago
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    one rule, if the bottom is a higher degree than the top, as x approaches infinity, the limit is zero

  18. amistre64
    • one year ago
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    if they are of the same degree, they limit to their leading coeffs .. its just simpler to divide off by the highest power of x and assess

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