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anonymous

  • 5 years ago

\[\lim x \rightarrow1 t+1 /(t-1)^2\]

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  1. anonymous
    • 5 years ago
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    \[\lim_{t \rightarrow 1}{t+1 \over (t-1)^2}=\infty\]

  2. anonymous
    • 5 years ago
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    yep! How'd you get it?

  3. anonymous
    • 5 years ago
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    Plugging \(t=1\) gives \(\frac{2}{0}\), which in undefined. Whenever you have a number other than \(0\),over \(0\) then the limit goes to \(\infty\) or\( - \infty\). All you have to do is to test the sign around \(1\), and you can see it's positive as x approaches 1 from both sides.

  4. anonymous
    • 5 years ago
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    I thought I'd have to factor it and get a real non-zero answer at first.

  5. anonymous
    • 5 years ago
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    when do you factor and try to get a non-zero denominator then?

  6. anonymous
    • 5 years ago
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    what?

  7. anonymous
    • 5 years ago
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    differentiate top and bottom :|

  8. anonymous
    • 5 years ago
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    you can't differentiate top and bottom

  9. anonymous
    • 5 years ago
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    its in the limits section, so not up to that yet.

  10. anonymous
    • 5 years ago
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    you can't apply l'hopital because it's not 0/0

  11. anonymous
    • 5 years ago
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    ohh yeh, twice and its zero lol

  12. anonymous
    • 5 years ago
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    You should try factoring when you have a \(\frac{0}{0}\) or a \(\frac{\infty}{0}\) or a similar form.

  13. anonymous
    • 5 years ago
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    oh -- thank you. I've been factoring whenever I get 0 in the denominator.

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spraguer (Moderator)
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