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anonymous

  • one year ago

What is the limit as x approaches infinity for f(x)=4x^2+8x/x^3+7x^2-x-9?

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  1. math&ing001
    • one year ago
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    When calculating limits near infinity we only take in consideration the highest degree term: \[\lim_{x \rightarrow \infty} \frac{ 4x ^{2}+8x }{ x ^{3}+7x ^{2}-x-9 }=\lim_{x \rightarrow \infty} \frac{ 4x ^{2}}{ x ^{3}}\]

  2. math&ing001
    • one year ago
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    @Camila1315 can you do it now ?

  3. anonymous
    • one year ago
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    Yes the limit would be infinity right?

  4. math&ing001
    • one year ago
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    Nope, how did you infinity ?

  5. math&ing001
    • one year ago
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    did you get*

  6. anonymous
    • one year ago
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    When I graphed it I saw the when x kept approaching infinity, the y value was coming from infinity.

  7. math&ing001
    • one year ago
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    No need to graph it, just go from here \(\lim_{x \rightarrow \infty} \frac{ 4x ^{2} }{ x ^{3} }\) and simplify.

  8. anonymous
    • one year ago
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    Oh is it 4?

  9. math&ing001
    • one year ago
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    \[\lim_{x \rightarrow \infty} \frac{ 4x ^{2} }{ x ^{3} }=\lim_{x \rightarrow \infty} \frac{ 4 }{ x }\] Do you know the limit of 1/x near infinity ?

  10. anonymous
    • one year ago
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    Im so bad with limits I thought the limit there would be infinity

  11. math&ing001
    • one year ago
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    O.K here's a graph that shows you how the 1/x function acts near infinity : |dw:1433595410014:dw|

  12. anonymous
    • one year ago
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    Oh so you have to look at x when it is approaching infinity?

  13. math&ing001
    • one year ago
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    Yeah, that's how you get the limit from the graph. Can you figure out the limit now ?

  14. anonymous
    • one year ago
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    Yeah so the limit would be 0

  15. math&ing001
    • one year ago
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    Yep !

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