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DLS

  • 2 years ago

Limits

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  1. DLS
    • 2 years ago
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    @mathslover

  2. mathslover
    • 2 years ago
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    It is easy , it just looks long. I will do that after lunch, ok?

  3. DLS
    • 2 years ago
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    \[\Huge \lim_{x \rightarrow 0}( \frac{1}{x} \int\limits_{0}^{a} e^{\sin^{2}t}dt-\int\limits_{x+y}^{a} e^{\sin^{2}t}dt)\]

  4. DLS
    • 2 years ago
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    Okay @mathslover and I know it is easy,I was trying it but i think the first term will cancel off because a and y both are constants maybe..i can be wrong,just drop a hint.

  5. DLS
    • 2 years ago
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    sorry the term 1/x is outside everything

  6. yrelhan4
    • 2 years ago
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    we can write the expression as.... (first term) + ∫ e^(sin^2 t dt) with limits ... a to x+y.. the sign changed on reversing the limits..

  7. yrelhan4
    • 2 years ago
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    now combining the two limits.. 0 to a and a to x+y .. 0 to x+y ..

  8. yrelhan4
    • 2 years ago
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    because the integrand is same..

  9. DLS
    • 2 years ago
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    0 to a and 0 to x+y=0 to x+y

  10. yrelhan4
    • 2 years ago
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    yep.

  11. DLS
    • 2 years ago
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    cool :O

  12. DLS
    • 2 years ago
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    \[\Huge \frac{1}{x} e^{\sin^{2}x+y}\]

  13. DLS
    • 2 years ago
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    differenatiate and bingo :O

  14. DLS
    • 2 years ago
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    e^sin^2x+y

  15. yrelhan4
    • 2 years ago
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    why did you not do anything to 1/x ?

  16. yrelhan4
    • 2 years ago
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    multiplication theorem.. hmm..

  17. DLS
    • 2 years ago
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    1/x was outside the integral sorry :P

  18. yrelhan4
    • 2 years ago
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    aur jab 0 put krenge.. to bhi to kuch aayega. vo kahan gyi? :P

  19. DLS
    • 2 years ago
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    Wait.Lets start again :|

  20. DLS
    • 2 years ago
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    \[\Huge \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{a}e^{\sin^{2}t} dt-\int\limits_{x+y}^{a}e^{\sin^{2}t} dt)\]

  21. DLS
    • 2 years ago
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    original question^

  22. DLS
    • 2 years ago
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    \[\Large \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{x+y}e^{\sin^{2}t} dt)\] On changing the sign and revevrsing the limits and combining them

  23. yrelhan4
    • 2 years ago
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    acha vo 'a' vala to 0 ho jayega.. a ki differentiation 0.. but hamne x+y ko differentiate bhi nhi kra.. :/

  24. DLS
    • 2 years ago
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    Apply LH Rule

  25. yrelhan4
    • 2 years ago
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    @shubhamsrg ?? :)

  26. shubhamsrg
    • 2 years ago
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    kya hua? kya dikkat hai ? (:

  27. DLS
    • 2 years ago
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    \[\Huge e^{\sin^{2}{x+y}}(1+\frac{dy}{dx})-e^{\sin^{2}y} \frac{dy}{dx}\]

  28. DLS
    • 2 years ago
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    put dy/dx=0

  29. yrelhan4
    • 2 years ago
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    why put dy/dx=0 ?

  30. DLS
    • 2 years ago
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    \[\Huge e^{\sin^{2}x+y}\] answer

  31. DLS
    • 2 years ago
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    y is behaving as a constant,we assume y independent of x

  32. DLS
    • 2 years ago
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    idk what i wrote please someone explain :P

  33. yrelhan4
    • 2 years ago
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    ohhh.. achaaaa..

  34. DLS
    • 2 years ago
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    If we are driving a car and chasing a thief on a bike then the rate of change of our speed won't affect the speed of the theif's bike

  35. yrelhan4
    • 2 years ago
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    limit y se x+y ?? how??

  36. DLS
    • 2 years ago
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    tune he to bataya?:/

  37. yrelhan4
    • 2 years ago
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    i said 0 to x+y.. :/

  38. DLS
    • 2 years ago
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    y->x+y hoga

  39. DLS
    • 2 years ago
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    wo "A" hai 0 nahi :|

  40. yrelhan4
    • 2 years ago
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    ohh. ok.

  41. DLS
    • 2 years ago
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    :)

  42. DLS
    • 2 years ago
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    but still didnt get it :|

  43. DLS
    • 2 years ago
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    \[\LARGE e^{\sin^{2}x+y}\] differentiating this^

  44. DLS
    • 2 years ago
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    \[\LARGE e^{\sin^{2}x+y} \times \sin2x(+0)\]

  45. DLS
    • 2 years ago
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    \[\LARGE e^{\sin^{2}y}\] Differentiating the lower limit^

  46. DLS
    • 2 years ago
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    \[\LARGE e^{\sin^{2}y}=>e^{\sin^{2}y} \times \sin2y\]

  47. DLS
    • 2 years ago
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    lower limit will become 0

  48. DLS
    • 2 years ago
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    I'm left with.. \[\LARGE \frac{1}{x}e^{\sin^{2}x+y} \times \sin2x\]

  49. yrelhan4
    • 2 years ago
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    abbe yr. x ki differentiation 1 hogi.. LH me numerator aur denominator alag alag diff krte hain.. :/

  50. DLS
    • 2 years ago
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    maine bhi to wahi bola tha BC

  51. DLS
    • 2 years ago
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    BUT 1 KA 0

  52. yrelhan4
    • 2 years ago
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    to x ki diff 1 ho jayegi.. aur numerator me tune jo kra tha. dy/dx=0 krke aa jayega ans.

  53. yrelhan4
    • 2 years ago
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    kya 1 ka 0? 1 baari diff krenge.. :/

  54. DLS
    • 2 years ago
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    1/x hai to 1 ka karenge to 0 x ka karenge to 1 0/1=0

  55. yrelhan4
    • 2 years ago
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    lol. 1*x ki diff. 0*x + 1*1.. lol

  56. yrelhan4
    • 2 years ago
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    to upar sirf 1 thodi hai.. baaki bhi to hai. usse multiply krega saara.. to 1* ....... = ......... aa jayega.. :O

  57. DLS
    • 2 years ago
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    badiya hai :|

  58. DLS
    • 2 years ago
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    par sin2x kahan gaya?

  59. DLS
    • 2 years ago
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    main question :P

  60. yrelhan4
    • 2 years ago
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    newton lebinitz me integrand ki differentiation nhi krte.. limits ko integrand me daalte hain multiply by limits ki diff..

  61. DLS
    • 2 years ago
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    medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal :*

  62. yrelhan4
    • 2 years ago
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    :')

  63. DLS
    • 2 years ago
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    seekho @shubhamsrg

  64. yrelhan4
    • 2 years ago
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    lol. meri bezti badi achi krte ho aap log. :/

  65. shubhamsrg
    • 2 years ago
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    maine upar kuch padha bhi nahi kya likha hai :3

  66. knock
    • 2 years ago
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    What's the answer then? xD

  67. shubhamsrg
    • 2 years ago
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    @yrelhan4 is the answer to all the questions . O:)

  68. yrelhan4
    • 2 years ago
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    :')

  69. knock
    • 2 years ago
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    Great xD I thought in a way to solve it but if it's already solved then ok (:

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