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haterofmath

  • 3 years ago

What hapeens if the limit is being taken from the left?

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  1. across
    • 3 years ago
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    Anything can happen, but generally, you're approaching the limit point from negative infinity.

  2. haterofmath
    • 3 years ago
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    so in a problem such as \[\lim_{x \rightarrow 2-}\frac{ x^2+4 }{ x-2 }\] howw would it work?

  3. helder_edwin
    • 3 years ago
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    since \(x\to2-\) then \(x<2\) so \(x-2<0\)

  4. helder_edwin
    • 3 years ago
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    this means that \[ \Large \lim_{x\to2-}\frac{x^2+4}{x-2}=-\infty \]

  5. haterofmath
    • 3 years ago
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    ok, so i won't have to plug anything in?

  6. haterofmath
    • 3 years ago
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    and if it were moving to the right it would be \[+\infty\]

  7. helder_edwin
    • 3 years ago
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    well, pluging in you would have \[ \Large \frac{2^2+4}{2-2}=\frac{8}{0} \]

  8. helder_edwin
    • 3 years ago
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    yes. if \(x\to2+\) then \(x>2\) so \(x-2>0\)

  9. pradipgr817
    • 3 years ago
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    I have tried it differently |dw:1347746473941:dw| at left as h tends 0 x tends\[2^{-}\]

  10. pradipgr817
    • 3 years ago
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    so if we replace x by 2-h and let h tends to 0 then there will be a different solution.

  11. haterofmath
    • 3 years ago
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    and what solutions that?

  12. pradipgr817
    • 3 years ago
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    you can put it and find it.

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