## DLS 2 years ago Limits

1. DLS

@mathslover

2. mathslover

It is easy , it just looks long. I will do that after lunch, ok?

3. DLS

$\Huge \lim_{x \rightarrow 0}( \frac{1}{x} \int\limits_{0}^{a} e^{\sin^{2}t}dt-\int\limits_{x+y}^{a} e^{\sin^{2}t}dt)$

4. DLS

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

sorry the term 1/x is outside everything

6. yrelhan4

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

now combining the two limits.. 0 to a and a to x+y .. 0 to x+y ..

8. yrelhan4

because the integrand is same..

9. DLS

0 to a and 0 to x+y=0 to x+y

10. yrelhan4

yep.

11. DLS

cool :O

12. DLS

$\Huge \frac{1}{x} e^{\sin^{2}x+y}$

13. DLS

differenatiate and bingo :O

14. DLS

e^sin^2x+y

15. yrelhan4

why did you not do anything to 1/x ?

16. yrelhan4

multiplication theorem.. hmm..

17. DLS

1/x was outside the integral sorry :P

18. yrelhan4

aur jab 0 put krenge.. to bhi to kuch aayega. vo kahan gyi? :P

19. DLS

Wait.Lets start again :|

20. DLS

$\Huge \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{a}e^{\sin^{2}t} dt-\int\limits_{x+y}^{a}e^{\sin^{2}t} dt)$

21. DLS

original question^

22. DLS

$\Large \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{x+y}e^{\sin^{2}t} dt)$ On changing the sign and revevrsing the limits and combining them

23. yrelhan4

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

24. DLS

Apply LH Rule

25. yrelhan4

@shubhamsrg ?? :)

26. shubhamsrg

kya hua? kya dikkat hai ? (:

27. DLS

$\Huge e^{\sin^{2}{x+y}}(1+\frac{dy}{dx})-e^{\sin^{2}y} \frac{dy}{dx}$

28. DLS

put dy/dx=0

29. yrelhan4

why put dy/dx=0 ?

30. DLS

$\Huge e^{\sin^{2}x+y}$ answer

31. DLS

y is behaving as a constant,we assume y independent of x

32. DLS

idk what i wrote please someone explain :P

33. yrelhan4

ohhh.. achaaaa..

34. DLS

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

limit y se x+y ?? how??

36. DLS

tune he to bataya?:/

37. yrelhan4

i said 0 to x+y.. :/

38. DLS

y->x+y hoga

39. DLS

wo "A" hai 0 nahi :|

40. yrelhan4

ohh. ok.

41. DLS

:)

42. DLS

but still didnt get it :|

43. DLS

$\LARGE e^{\sin^{2}x+y}$ differentiating this^

44. DLS

$\LARGE e^{\sin^{2}x+y} \times \sin2x(+0)$

45. DLS

$\LARGE e^{\sin^{2}y}$ Differentiating the lower limit^

46. DLS

$\LARGE e^{\sin^{2}y}=>e^{\sin^{2}y} \times \sin2y$

47. DLS

lower limit will become 0

48. DLS

I'm left with.. $\LARGE \frac{1}{x}e^{\sin^{2}x+y} \times \sin2x$

49. yrelhan4

abbe yr. x ki differentiation 1 hogi.. LH me numerator aur denominator alag alag diff krte hain.. :/

50. DLS

maine bhi to wahi bola tha BC

51. DLS

BUT 1 KA 0

52. yrelhan4

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

53. yrelhan4

kya 1 ka 0? 1 baari diff krenge.. :/

54. DLS

1/x hai to 1 ka karenge to 0 x ka karenge to 1 0/1=0

55. yrelhan4

lol. 1*x ki diff. 0*x + 1*1.. lol

56. yrelhan4

to upar sirf 1 thodi hai.. baaki bhi to hai. usse multiply krega saara.. to 1* ....... = ......... aa jayega.. :O

57. DLS

58. DLS

par sin2x kahan gaya?

59. DLS

main question :P

60. yrelhan4

newton lebinitz me integrand ki differentiation nhi krte.. limits ko integrand me daalte hain multiply by limits ki diff..

61. DLS

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

:')

63. DLS

seekho @shubhamsrg

64. yrelhan4

lol. meri bezti badi achi krte ho aap log. :/

65. shubhamsrg

maine upar kuch padha bhi nahi kya likha hai :3

66. knock

67. shubhamsrg

@yrelhan4 is the answer to all the questions . O:)

68. yrelhan4

:')

69. knock

Great xD I thought in a way to solve it but if it's already solved then ok (: