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lim-x-0-e-sin-x-e-x-sin-x-x-




Question Number 211759 by liuxinnan last updated on 20/Sep/24
lim_(x→0) ((e^(sin x) −e^x )/(sin x−x))=?
limx0esinxexsinxx=?
Commented by liuxinnan last updated on 20/Sep/24
Answered by TonyCWX08 last updated on 20/Sep/24
It′s 1
Its1
Answered by TonyCWX08 last updated on 20/Sep/24
Using L′Hopital rule  Find the derivative for the numerator and denominator:  (d/dx)(e^(sin(x)) −e^x )=cos(x)e^(sin(x)) −e^x   (d/dx)(sin x − x )=cos (x)−1    The limit now transform to  lim_(x→0) (((cos(x)e^(sin(x)) −e^x )/(cos (x)−1)))  lim_(x→0) (((cos(x)e^(sin(x)) −e^x )/(cos (x)−1)))    But it′s still not defined, so we apply the rule again.  (d/dx)(cos(x)e^(sin(x)) −e^x )=−sin(x)e^(sin(x)) +cos^2 (x)e^(sin(x)) −e^x   (d/dx)(cos(x)−1)=−sin(x)    The limit now transform to  lim_(x→0) (((−sin(x)e^(sin(x)) +cos^2 (x)e^(sin(x)) −e^x )/(−sin(x))))  =lim_(x→0) (((sin(x)e^(sin(x)) −cos^2 (x)e^(sin(x)) +e^x )/(sin(x))))    Still undefined, apply rule again  (d/dx)(sin(x)e^(sin(x)) −cos^2 (x)e^(sin(x)) +e^x )=−e^(sin(x)) cos(x)sin^2 (x)−e^(sin(x)) sin(2x)−e^(sin(x)) cos(x)sin(x)−e^x   (d/dx)(sin(x))=cos(x)    The limit now transform to  lim_(x→0) (((−e^(sin(x)) cos(x)sin^2 (x)−e^(sin(x)) sin(2x)−e^(sin(x)) cos(x)sin(x)−e^x )/(cos(x)))  Finally, substitute x=0  The limit is 1
UsingLHopitalruleFindthederivativeforthenumeratoranddenominator:ddx(esin(x)ex)=cos(x)esin(x)exddx(sinxx)=cos(x)1Thelimitnowtransformtolimx0(cos(x)esin(x)excos(x)1)limx0(cos(x)esin(x)excos(x)1)Butitsstillnotdefined,soweapplytheruleagain.ddx(cos(x)esin(x)ex)=sin(x)esin(x)+cos2(x)esin(x)exddx(cos(x)1)=sin(x)Thelimitnowtransformtolimx0(sin(x)esin(x)+cos2(x)esin(x)exsin(x))=limx0(sin(x)esin(x)cos2(x)esin(x)+exsin(x))Stillundefined,applyruleagainddx(sin(x)esin(x)cos2(x)esin(x)+ex)=esin(x)cos(x)sin2(x)esin(x)sin(2x)esin(x)cos(x)sin(x)exddx(sin(x))=cos(x)Thelimitnowtransformtolimx0(esin(x)cos(x)sin2(x)esin(x)sin(2x)esin(x)cos(x)sin(x)excos(x)Finally,substitutex=0Thelimitis1
Answered by BHOOPENDRA last updated on 20/Sep/24
As we know  lim_(x→0)  ((e^x −1)/(x ))=1  so from there   lim_(x→0)  e^x     lim_(x→0)  (((e^(sinx−x) −1)/(sinx−x)))     ⇒ e^0  ×1  =1
Asweknowlimx0ex1x=1sofromtherelimx0exlimx0(esinxx1sinxx)e0×1=1
Answered by mehdee7396 last updated on 20/Sep/24
e^(sinx) =1+x+(x^2 /2)+O(x^4 )  e^x =1+x+(x^2 /2)+(x^3 /(3!))+O(x^4 )  ⇒lim_(x→0)  ((e^(sinx) −e^x )/(sinx−x))=lim_(x→0)  ((−(1/6)x^3 )/(((−1)/6)x^3 )) =1 ✓
esinx=1+x+x22+O(x4)ex=1+x+x22+x33!+O(x4)limx0esinxexsinxx=limx016x316x3=1
Commented by liuxinnan last updated on 20/Sep/24
Thanks everyone.It helps me a lot.
Thankseveryone.Ithelpsmealot.

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