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DLS
 3 years ago
Limits
DLS
 3 years ago
Limits

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mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0It is easy , it just looks long. I will do that after lunch, ok?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Huge \lim_{x \rightarrow 0}( \frac{1}{x} \int\limits_{0}^{a} e^{\sin^{2}t}dt\int\limits_{x+y}^{a} e^{\sin^{2}t}dt)\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1Okay @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
 3 years ago
Best ResponseYou've already chosen the best response.1sorry the term 1/x is outside everything

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2we 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..

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2now combining the two limits.. 0 to a and a to x+y .. 0 to x+y ..

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2because the integrand is same..

DLS
 3 years ago
Best ResponseYou've already chosen the best response.10 to a and 0 to x+y=0 to x+y

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Huge \frac{1}{x} e^{\sin^{2}x+y}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1differenatiate and bingo :O

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2why did you not do anything to 1/x ?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2multiplication theorem.. hmm..

DLS
 3 years ago
Best ResponseYou've already chosen the best response.11/x was outside the integral sorry :P

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2aur jab 0 put krenge.. to bhi to kuch aayega. vo kahan gyi? :P

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Huge \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{a}e^{\sin^{2}t} dt\int\limits_{x+y}^{a}e^{\sin^{2}t} dt)\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Large \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{x+y}e^{\sin^{2}t} dt)\] On changing the sign and revevrsing the limits and combining them

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2acha vo 'a' vala to 0 ho jayega.. a ki differentiation 0.. but hamne x+y ko differentiate bhi nhi kra.. :/

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0kya hua? kya dikkat hai ? (:

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Huge e^{\sin^{2}{x+y}}(1+\frac{dy}{dx})e^{\sin^{2}y} \frac{dy}{dx}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Huge e^{\sin^{2}x+y}\] answer

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1y is behaving as a constant,we assume y independent of x

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1idk what i wrote please someone explain :P

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1If 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

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2limit y se x+y ?? how??

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE e^{\sin^{2}x+y}\] differentiating this^

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE e^{\sin^{2}x+y} \times \sin2x(+0)\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE e^{\sin^{2}y}\] Differentiating the lower limit^

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE e^{\sin^{2}y}=>e^{\sin^{2}y} \times \sin2y\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1I'm left with.. \[\LARGE \frac{1}{x}e^{\sin^{2}x+y} \times \sin2x\]

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2abbe yr. x ki differentiation 1 hogi.. LH me numerator aur denominator alag alag diff krte hain.. :/

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1maine bhi to wahi bola tha BC

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2to x ki diff 1 ho jayegi.. aur numerator me tune jo kra tha. dy/dx=0 krke aa jayega ans.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2kya 1 ka 0? 1 baari diff krenge.. :/

DLS
 3 years ago
Best ResponseYou've already chosen the best response.11/x hai to 1 ka karenge to 0 x ka karenge to 1 0/1=0

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2lol. 1*x ki diff. 0*x + 1*1.. lol

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2to upar sirf 1 thodi hai.. baaki bhi to hai. usse multiply krega saara.. to 1* ....... = ......... aa jayega.. :O

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2newton lebinitz me integrand ki differentiation nhi krte.. limits ko integrand me daalte hain multiply by limits ki diff..

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1medal 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 :*

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2lol. meri bezti badi achi krte ho aap log. :/

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0maine upar kuch padha bhi nahi kya likha hai :3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What's the answer then? xD

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0@yrelhan4 is the answer to all the questions . O:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Great xD I thought in a way to solve it but if it's already solved then ok (:
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