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anonymous

  • one year ago

State the value of the limit, if it exists lim x→0 sin(x) multiplied with 3x^3+2x^2/x^2 Anybody? :)

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  1. anonymous
    • one year ago
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    ok this is a big one

  2. anonymous
    • one year ago
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    so stay with me

  3. anonymous
    • one year ago
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    first we separate the lim

  4. anonymous
    • one year ago
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    \[\lim_{x \rightarrow 0} (3x+2) \times \lim_{x \rightarrow 0}(\frac{ \sin(x) }{ x ^{2} })\]

  5. anonymous
    • one year ago
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    i forgot to times x^2 with sin(x)

  6. anonymous
    • one year ago
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    ok

  7. anonymous
    • one year ago
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    \[\lim_{x \rightarrow 0}(\frac{ \sin(x)x ^{2} }{ x ^{2} }) \times \lim_{x \rightarrow 0} (3x+2)\]

  8. anonymous
    • one year ago
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    ur following the steps right

  9. anonymous
    • one year ago
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    wow! :)

  10. anonymous
    • one year ago
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    Yep! Think so…

  11. anonymous
    • one year ago
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    3x+2 is a polynomial and thus everywhere continuous so \[\lim_{x \rightarrow 0} (3x+2)= 2+3(0)=2\]

  12. anonymous
    • one year ago
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    and the \[\lim_{x \rightarrow 0} \frac{ \sin(0) \times 0 }{ 0 } =0\]

  13. anonymous
    • one year ago
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    \[ 2 \times 0=0\]

  14. anonymous
    • one year ago
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    got it

  15. anonymous
    • one year ago
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    if u have similar denominator like x^2 in ur case u can separate the limit

  16. anonymous
    • one year ago
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    Yup! Think so…. :)

  17. anonymous
    • one year ago
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    I have to go through it a few times to make sure I understand everything :)

  18. anonymous
    • one year ago
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    Thank you so much!

  19. anonymous
    • one year ago
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    Any possibilities of some follow-up questions later on?

  20. anonymous
    • one year ago
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    Yea sure

  21. IrishBoy123
    • one year ago
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    you simply can not say this: \[\lim_{x \rightarrow 0}(\frac{ \sin(x)x ^{2} }{ x ^{2} }) \] \[\implies \lim_{x \rightarrow 0} \frac{ \sin(0) \times 0 }{ 0 } =0\] the whole point of this is that dividing by zero you cannot do, so you check what the function in question does as you approach very close to zero. so: \[\lim\limits_{ x→0} \, \, \sin(x) .\frac{3x^3+2x^2}{x^2}\] \[=\lim\limits_{ x→0} \, \, \sin(x) .\lim\limits_{ x→0}\frac{3x^3+2x^2}{x^2}\] \[=\lim\limits_{ x→0} \, \, \sin(x) .\lim\limits_{ x→0}\frac{\frac{3x^3}{x^2}+\frac{2x^2}{x^2}}{\frac{x^2}{x^2}}\] \[=\lim\limits_{ x→0} \, \, \sin(x) .\lim\limits_{ x→0}\frac{3x+2}{1}\] \[=0 \times 2 = 0\]

  22. anonymous
    • one year ago
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    Thank you so much for your reply! :)

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