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  • DLS

Limits

Mathematics
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  • DLS
It is easy , it just looks long. I will do that after lunch, ok?
  • DLS
\[\Huge \lim_{x \rightarrow 0}( \frac{1}{x} \int\limits_{0}^{a} e^{\sin^{2}t}dt-\int\limits_{x+y}^{a} e^{\sin^{2}t}dt)\]

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Other answers:

  • 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.
  • DLS
sorry the term 1/x is outside everything
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..
now combining the two limits.. 0 to a and a to x+y .. 0 to x+y ..
because the integrand is same..
  • DLS
0 to a and 0 to x+y=0 to x+y
yep.
  • DLS
cool :O
  • DLS
\[\Huge \frac{1}{x} e^{\sin^{2}x+y}\]
  • DLS
differenatiate and bingo :O
  • DLS
e^sin^2x+y
why did you not do anything to 1/x ?
multiplication theorem.. hmm..
  • DLS
1/x was outside the integral sorry :P
aur jab 0 put krenge.. to bhi to kuch aayega. vo kahan gyi? :P
  • DLS
Wait.Lets start again :|
  • DLS
\[\Huge \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{a}e^{\sin^{2}t} dt-\int\limits_{x+y}^{a}e^{\sin^{2}t} dt)\]
  • DLS
original question^
  • 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
acha vo 'a' vala to 0 ho jayega.. a ki differentiation 0.. but hamne x+y ko differentiate bhi nhi kra.. :/
  • DLS
Apply LH Rule
kya hua? kya dikkat hai ? (:
  • DLS
\[\Huge e^{\sin^{2}{x+y}}(1+\frac{dy}{dx})-e^{\sin^{2}y} \frac{dy}{dx}\]
  • DLS
put dy/dx=0
why put dy/dx=0 ?
  • DLS
\[\Huge e^{\sin^{2}x+y}\] answer
  • DLS
y is behaving as a constant,we assume y independent of x
  • DLS
idk what i wrote please someone explain :P
ohhh.. achaaaa..
  • 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
limit y se x+y ?? how??
  • DLS
tune he to bataya?:/
i said 0 to x+y.. :/
  • DLS
y->x+y hoga
  • DLS
wo "A" hai 0 nahi :|
ohh. ok.
  • DLS
:)
  • DLS
but still didnt get it :|
  • DLS
\[\LARGE e^{\sin^{2}x+y}\] differentiating this^
  • DLS
\[\LARGE e^{\sin^{2}x+y} \times \sin2x(+0)\]
  • DLS
\[\LARGE e^{\sin^{2}y}\] Differentiating the lower limit^
  • DLS
\[\LARGE e^{\sin^{2}y}=>e^{\sin^{2}y} \times \sin2y\]
  • DLS
lower limit will become 0
  • DLS
I'm left with.. \[\LARGE \frac{1}{x}e^{\sin^{2}x+y} \times \sin2x\]
abbe yr. x ki differentiation 1 hogi.. LH me numerator aur denominator alag alag diff krte hain.. :/
  • DLS
maine bhi to wahi bola tha BC
  • DLS
BUT 1 KA 0
to x ki diff 1 ho jayegi.. aur numerator me tune jo kra tha. dy/dx=0 krke aa jayega ans.
kya 1 ka 0? 1 baari diff krenge.. :/
  • DLS
1/x hai to 1 ka karenge to 0 x ka karenge to 1 0/1=0
lol. 1*x ki diff. 0*x + 1*1.. lol
to upar sirf 1 thodi hai.. baaki bhi to hai. usse multiply krega saara.. to 1* ....... = ......... aa jayega.. :O
  • DLS
badiya hai :|
  • DLS
par sin2x kahan gaya?
  • DLS
main question :P
newton lebinitz me integrand ki differentiation nhi krte.. limits ko integrand me daalte hain multiply by limits ki diff..
  • 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 :*
:')
  • DLS
lol. meri bezti badi achi krte ho aap log. :/
maine upar kuch padha bhi nahi kya likha hai :3
What's the answer then? xD
@yrelhan4 is the answer to all the questions . O:)
:')
Great xD I thought in a way to solve it but if it's already solved then ok (:

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