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Question Number 80670 by jagoll last updated on 05/Feb/20

$$\underset{x\to \pi }{\lim }\frac{e^{\sin x}-1}{x-\pi }=?$$

Commented by jagoll last updated on 05/Feb/20

$$\underset{x\to \pi }{\lim }\frac{\cos x.e^{\sin x}}{1}=-1$$

Commented by abdomathmax last updated on 05/Feb/20

$$letf\left(x\right)=\frac{e^{sinx}-1}{x-\pi }changementx-\pi =tgive$$$$f\left(x\right)=g\left(t\right)=\frac{e^{sin(\pi +t)}-1}{t}=\frac{e^{-sint}-1}{t}$$$$\sim \frac{e^{-t}-1}{t}\sim \frac{1-t-1}{t}=-1\Rightarrow {lim}_{x\to \pi }f\left(x\right)=-1$$

Answered by $@ty@m123 last updated on 05/Feb/20

$$Letx-\pi =t$$$$x=\pi +t$$$$Asx\to \pi ,t=0$$$$\underset{t\to 0}{\lim }\frac{e^{\sin (\pi +t)}-1}{t}$$$$\underset{t\to 0}{\lim }\frac{e^{-\sin t}-1}{t}$$$$\underset{t\to 0}{\lim }\frac{1-\sin t+\frac{\sin {}^{2}t}{2!}+.....-1}{t}$$$$\underset{t\to 0}{\lim }\frac{-\sin t}{t}$$$$=-1$$

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