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nubeer

ax^2+bx-2=0 have 2 roots alpha and beta and question is. limx-->beta [(1-cos(ax^2+bx-2)/((ax^2+bx-2)^2)

  • 2 years ago
  • 2 years ago

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  1. myininaya
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    if x goes to beta, then ax^2+bx-2 goes to 0 since beta is a root of ax^2+bx-2 so thats means we have 0/0 when we plug in beta we could use l'hospital's rule here \[\lim_{x \rightarrow \beta}\frac{0-(-\sin(ax^2+bx-2))(2ax+b)}{2(ax^2+bx-2)(2ax+b)}\]

    • 2 years ago
  2. myininaya
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    \[\lim_{x \rightarrow \beta}\frac{\sin(ax^2+bx-2)}{2(ax^2+bx-2)}\]

    • 2 years ago
  3. myininaya
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    but when we plug in beta we still have 0/0

    • 2 years ago
  4. myininaya
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    so we can use l'hosptial again

    • 2 years ago
  5. myininaya
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    \[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx-2)(2ax+b)}{2(2ax+b)}\]

    • 2 years ago
  6. myininaya
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    \[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx-2)}{2}=\frac{\cos(0)}{2}=\frac{1}{2}\]

    • 2 years ago
  7. nubeer
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    wow dude ur answer is right but still have to think how u reach here

    • 2 years ago
  8. myininaya
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    do you know l'hospital?

    • 2 years ago
  9. myininaya
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    however you spell that dead math guy's name

    • 2 years ago
  10. nubeer
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    hmmm ya i heard of it but never used it

    • 2 years ago
  11. myininaya
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    so let me show you another way one sec

    • 2 years ago
  12. nubeer
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    ok but i still ddnt get what ever u answer because of that sinx

    • 2 years ago
  13. myininaya
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    let \[V(x)=ax^2+bx-2\] we are given \[V(\beta)=0\] so if x->beta, then V->0 so we have \[\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2}\] \[=\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2} \cdot \frac{1+\cos(V)}{1+\cos(V)}\] \[=\lim_{v \rightarrow 0}\frac{1-\cos^2(V)}{V^2(1+\cos(V)}=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2(1+\cos(V))}\] \[=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2} \cdot \frac{1}{1+\cos(V))}\] \[=1^2 \cdot \frac{1}{1+\cos(0)}=\frac{1}{1+1}=\frac{1}{2}\]

    • 2 years ago
  14. nubeer
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    thanks man this thing really helped me out

    • 2 years ago
  15. myininaya
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    thanks girl* :)

    • 2 years ago
  16. nubeer
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    lol i am a boy. :P

    • 2 years ago
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