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DLS

  • 2 years ago

limits(38)

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  1. hba
    • 2 years ago
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    lol wut?

  2. DLS
    • 2 years ago
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    Let \(\alpha\) and \(\beta\) be the distinct roots of \(\ ax^2+bx+c=0\) then \[\Huge \lim_{x \rightarrow \alpha} \frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2}=?\]

  3. DLS
    • 2 years ago
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    Attempt: \[\Large \frac{1-cosa(x-\alpha)(x-\beta)}{(x-\alpha)^2}\]

  4. DLS
    • 2 years ago
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    \[\lim_{x \rightarrow \alpha}\Large \frac{1-cosa(x-\beta)}{(x-\alpha)}\]

  5. yrelhan4
    • 2 years ago
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    1-cos(2x)=2 sin^2 x laga.. quadratic ko khol mat.

  6. DLS
    • 2 years ago
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    without opening the quadratic it wont solve :/ (x-alpha)^2 wont cancel i guess

  7. yrelhan4
    • 2 years ago
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    do it.. sin^2 something aayega.. (x-alpha)^2 hai.. so [ sin(something)/(x-alpha)] ^2 ho jayega..

  8. DLS
    • 2 years ago
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    \[\Large \frac{ 2\sin^{2} (ax^2+bx+c)}{(x-\alpha)^2}\] this?

  9. DLS
    • 2 years ago
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    sorry upon 2 too

  10. DLS
    • 2 years ago
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    \[\Large \frac{ 2\sin^{2} \frac{ (ax^2+bx+c)}{2}}{(x-\alpha)^2}\]

  11. yrelhan4
    • 2 years ago
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    yup. ab quadratic ko (x-alpha)(x-beta) likh.. multiply divide by (x-beta)^2 ka whole square.. [sin(x-alpha)(x-beta)/(x-alpha)(x-beta)]^2 *2(x-beta)^2.. pehli term sinx/x ke form me hai to 1.. bach gya 2(x-beta)^2.. to aa gya. 2*(alpha-beta)^2 ??

  12. DLS
    • 2 years ago
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    answer to galat hai khair :P

  13. yrelhan4
    • 2 years ago
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    pata nhi..

  14. DLS
    • 2 years ago
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    \[(a)0\] \[(b) \frac{1}{2}a^(\alpha-\beta)^2\] \[(c) \frac{1}{2}^(\alpha-\beta)^2\] \[(d) \frac{-1}{2}a^2(\alpha-\beta)^2\]

  15. yrelhan4
    • 2 years ago
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    ohho.. ax^2 + bx + c/2 tha.. :/

  16. yrelhan4
    • 2 years ago
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    C hai?

  17. DLS
    • 2 years ago
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    nah :P

  18. yrelhan4
    • 2 years ago
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    pta nhi.. aapko kya lgta hai?

  19. DLS
    • 2 years ago
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    x-beta wala step (2nd step) samaj nhi aya

  20. yrelhan4
    • 2 years ago
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    (x-beta)^2 se multiply divide kr diya..

  21. DLS
    • 2 years ago
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    kyunkiya?

  22. yrelhan4
    • 2 years ago
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    taaki sinx/x vali form ban sake.. [sin(x-alpha)(x-beta)/2]^2 hai.. to niche [(x-alpha)(x-beta)/2]^2 chahiye.. to (x-beta)^2/4 se multiply krenge..

  23. DLS
    • 2 years ago
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    bus ab rukja :P

  24. DLS
    • 2 years ago
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    @yrelhan4 (x-alpha) nahi use kar sakte (x-beta) ki jagah?

  25. yrelhan4
    • 2 years ago
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    kaise? sin me x-alpha aur x- beta dono hain na.. to dono lane pdenge denominator me..

  26. DLS
    • 2 years ago
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    OH :O ab bus rukja :P

  27. DLS
    • 2 years ago
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    yo aagaya im the winner :P B hai

  28. DLS
    • 2 years ago
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    \[\LARGE \frac{2(\sin \frac{(a(x-\alpha)(x-\beta)}{2})^2}{(x-\alpha)^2} \times \frac{(\frac{a}{2})^2 \times (x-\beta)^2}{ (\frac{a}{2})^2 \times (x-\beta)^2}\]

  29. DLS
    • 2 years ago
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    u forgot about a :P

  30. yrelhan4
    • 2 years ago
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    ah.. samajh gya. :/

  31. DLS
    • 2 years ago
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    :D

  32. yrelhan4
    • 2 years ago
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    hmm iit. hmm

  33. DLS
    • 2 years ago
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    :P

  34. DLS
    • 2 years ago
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    1 ghanta ek question me lag raha hai :| ye 5minute ka tha,shubham ko ek minute me strike kar jata :|

  35. yrelhan4
    • 2 years ago
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    ok bye.

  36. DLS
    • 2 years ago
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    thanks for the help :)

  37. yrelhan4
    • 2 years ago
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    haha. :')

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