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clara1223

  • one year ago

find the limit as h approaches 0 of h(3+(2/h^2))

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  1. clara1223
    • one year ago
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    \[\lim_{h \rightarrow 0} h(3+\frac{ 2 }{ h ^{2} })\]

  2. amistre64
    • one year ago
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    we can try some test points, say h=.0000001, and h=-.0000001 but what does your thoughts tell you initially?

  3. clara1223
    • one year ago
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    Well I solved and got that the limit is 2, but that was wrong.

  4. amistre64
    • one year ago
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    your solution was flawed then :) we have 2 terms: ab if h < 0, a=-, b=+ ... ab = - if h > 0, a=+,b=+ ... ab=+

  5. clara1223
    • one year ago
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    I don't understand.

  6. amistre64
    • one year ago
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    let a=h let b=3+2/h^2 then multiply them together and yo have your setup. when h is less than zero, what are our signs for a and b? what about their product?

  7. clara1223
    • one year ago
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    if h is less than 0 then a would be negative and b would be positive, and their products would both be negative

  8. amistre64
    • one year ago
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    ok, so we have a left side limit that is negative ... now assume h is greater than zero ... what are the signs of a and b? and of their product?

  9. clara1223
    • one year ago
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    if h is greater than 0 then a and b would both be positive and so would be their products.

  10. clara1223
    • one year ago
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    oh, so since from the left it's negative and from the right it's positive, it doesnt exist?

  11. amistre64
    • one year ago
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    so we have a right side limit that is positive can a positive limit ever be equal to a negative limit?

  12. amistre64
    • one year ago
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    correct, its actually -inf and +inf so yeah

  13. clara1223
    • one year ago
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    yeah, I graphed it and it helped me to understand.

  14. amistre64
    • one year ago
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    graphs do help if applicable :)

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