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Question Number 211759 by liuxinnan last updated on 20/Sep/24

$$\underset{x\to 0}{\lim }\frac{e^{\sin x}-e^x}{\sin x-x}=?$$

Commented by liuxinnan last updated on 20/Sep/24

Answered by TonyCWX08 last updated on 20/Sep/24

$${It}^{\prime }s1$$

Answered by TonyCWX08 last updated on 20/Sep/24

$$Using{L}^{\prime }Hopitalrule$$$$Findthederivativeforthenumeratoranddenominator:$$$$\frac{d}{dx}(e^{\sin \left(x\right)}-e^x)=\cos \left(x\right)e^{\sin \left(x\right)}-e^x$$$$\frac{d}{dx}(\sin x-x)=\cos \left(x\right)-1$$$$$$$$Thelimitnowtransformto$$$$\underset{x\to 0}{\lim }\left(\frac{cos\left(x\right)e^{\sin \left(x\right)}-e^x}{\cos \left(x\right)-1}\right)$$$$\underset{x\to 0}{\lim }\left(\frac{cos\left(x\right)e^{\sin \left(x\right)}-e^x}{\cos \left(x\right)-1}\right)$$$$$$$$But{it}^{\prime }sstillnotdefined,soweapplytheruleagain.$$$$\frac{d}{dx}(cos\left(x\right)e^{sin\left(x\right)}-e^x)=-sin\left(x\right)e^{sin\left(x\right)}+{cos}^2\left(x\right)e^{sin\left(x\right)}-e^x$$$$\frac{d}{dx}(cos\left(x\right)-1)=-sin\left(x\right)$$$$$$$$Thelimitnowtransformto$$$$\underset{x\to 0}{\lim }\left(\frac{-sin\left(x\right)e^{sin\left(x\right)}+{cos}^2\left(x\right)e^{sin\left(x\right)}-e^x}{-sin\left(x\right)}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{sin\left(x\right)e^{sin\left(x\right)}-{cos}^2\left(x\right)e^{sin\left(x\right)}+e^x}{sin\left(x\right)}\right)$$$$$$$$Stillundefined,applyruleagain$$$$\frac{d}{dx}(sin\left(x\right)e^{sin\left(x\right)}-{cos}^2\left(x\right)e^{sin\left(x\right)}+e^x)=-e^{sin\left(x\right)}cos\left(x\right){sin}^2\left(x\right)-e^{sin\left(x\right)}sin\left(2x\right)-e^{sin\left(x\right)}cos\left(x\right)sin\left(x\right)-e^x$$$$\frac{d}{dx}\left(sin\left(x\right)\right)=cos\left(x\right)$$$$$$$$Thelimitnowtransformto$$$$\underset{x\to 0}{\lim }(\frac{-e^{sin\left(x\right)}cos\left(x\right){sin}^2\left(x\right)-e^{sin\left(x\right)}sin\left(2x\right)-e^{sin\left(x\right)}cos\left(x\right)sin\left(x\right)-e^x}{cos\left(x\right)}$$$$Finally,substitutex=0$$$$Thelimitis1$$

Answered by BHOOPENDRA last updated on 20/Sep/24

$$Asweknow\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$$$$sofromthere$$$$\underset{x\to 0}{\lim }{e}^x\underset{x\to 0}{\lim }\left(\frac{e^{sinx-x}-1}{sinx-x}\right)$$$$\Rightarrow e^0\times 1$$$$=1$$

Answered by mehdee7396 last updated on 20/Sep/24

$$e^{sinx}=1+x+\frac{x^2}{2}+O\left(x^4\right)$$$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+O\left(x^4\right)$$$$\Rightarrow {lim}_{x\to 0}\frac{e^{sinx}-e^x}{sinx-x}={lim}_{x\to 0}\frac{-\frac{1}{6}{x}^3}{\frac{-1}{6}{x}^3}=1✓$$

Commented by liuxinnan last updated on 20/Sep/24

$$Thankseveryone.Ithelpsmealot.$$

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