Question

Limits

Answer 1

@mathslover

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

sorry the term 1/x is outside everything

Answer 6

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..

Answer 7

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

Answer 8

because the integrand is same..

Answer 9

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

Answer 10

yep.

Answer 11

cool :O

Answer 12

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

Answer 13

differenatiate and bingo :O

Answer 14

e^sin^2x+y

Answer 15

why did you not do anything to 1/x ?

Answer 16

multiplication theorem.. hmm..

Answer 17

1/x was outside the integral sorry :P

Answer 18

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

Answer 19

Wait.Lets start again :|

Answer 20

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

Answer 21

original question^

Answer 22

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

Answer 23

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

Answer 24

Apply LH Rule

Answer 25

@shubhamsrg ?? :)

Answer 26

kya hua? kya dikkat hai ? (:

Answer 27

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

Answer 28

put dy/dx=0

Answer 29

why put dy/dx=0 ?

Answer 30

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

Answer 31

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

Answer 32

idk what i wrote please someone explain :P

Answer 33

ohhh.. achaaaa..

Answer 34

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

Answer 35

limit y se x+y ?? how??

Answer 36

tune he to bataya?:/

Answer 37

i said 0 to x+y.. :/

Answer 38

y->x+y hoga

Answer 39

wo "A" hai 0 nahi :|

Answer 40

ohh. ok.

Answer 41

:)

Answer 42

but still didnt get it :|

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

lower limit will become 0

Answer 48

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

Answer 49

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

Answer 50

maine bhi to wahi bola tha BC

Answer 51

BUT 1 KA 0

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

badiya hai :|

Answer 58

par sin2x kahan gaya?

Answer 59

main question :P

Answer 60

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

Answer 61

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 :*

Answer 62

:')

Answer 63

seekho @shubhamsrg

Answer 64

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

Answer 65

maine upar kuch padha bhi nahi kya likha hai :3

Answer 66

What's the answer then? xD

Answer 67

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

Answer 68

:')

Answer 69

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