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anonymous

  • one year ago

What is the limit of (x)÷((1/1+x)−1) as x approaches 0?

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  1. anonymous
    • one year ago
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    \[\large \lim_{x \rightarrow 0} \frac{ x }{ \left( \frac{ 1 }{ 1+x }-1 \right) }\] this?

  2. anonymous
    • one year ago
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    yes!

  3. anonymous
    • one year ago
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    Lets start off by simplifying the denominator you can use the following \[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

  4. anonymous
    • one year ago
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    Go ahead and see what you get

  5. anonymous
    • one year ago
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    So then would the denominator be \[(1-1+x)/(1+x)\]

  6. anonymous
    • one year ago
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    wait, the first part should be 1-1-x I think.

  7. anonymous
    • one year ago
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    \[\huge \lim_{x \rightarrow 0} \frac{ x }{ \left( \frac{ 1-1(1+x) }{ 1+x } \right) } = \lim_{x \rightarrow 0}\frac{ x }{ \left( \frac{ 1-1-x }{ 1+x } \right) }\] correct?

  8. anonymous
    • one year ago
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    Keep simplifying :)

  9. anonymous
    • one year ago
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    Yeah. And then I just use the reciprocal of the denominator and plug in the 0, and I get -1?

  10. anonymous
    • one year ago
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    Bingo!

  11. anonymous
    • one year ago
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    thank you!!

  12. anonymous
    • one year ago
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    Yw :)

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