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Question Number 81096 by jagoll last updated on 09/Feb/20

$$\underset{x\to 3}{\lim }{\left(\frac{\sin x}{\sin 3}\right)}^{\frac{1}{x-3}}=?$$

Commented by abdomathmax last updated on 09/Feb/20

$$letf\left(x\right)={\left(\frac{sinx}{sin3}\right)}^{\frac{1}{x-3}}changementx-3=tgive$$$$f\left(x\right)=g\left(t\right)={\left(\frac{sin(t+3)}{sin3}\right)}^{\frac{1}{t}}={\left(\frac{sintcos3+costsin3}{sin3}\right)}^{\frac{1}{t}}$$$$={(cotan3sint+cost)}^{\frac{1}{t}}$$$$=e^{\frac{1}{t}ln(cost+cotan3sint)}$$$$cost\sim 1-\frac{t^2}{2}andsint\sim t\Rightarrow$$$$cost+cotan\left(3\right)sint\sim 1-\frac{t^2}{2}+cotan\left(3\right)t\Rightarrow$$$$ln(...)\sim cotan\left(3\right)t-\frac{t^2}{2}\Rightarrow$$$$\frac{1}{t}ln(cost+cotan\left(3\right))sint)\sim cotan3-\frac{t}{2}\to cotan3$$$$(t\to 0)\Rightarrow {lim}_{t\to 0}g\left(t\right)=e^{cotan3}=e^{\frac{cos3}{sin3}}={lim}_{x\to 3}f\left(x\right)$$

Answered by john santu last updated on 09/Feb/20

$$limitform{\left(1\right)}^{\mathrm{\infty }}$$$$weknowthat\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x=e$$$$letx-3=t\Rightarrow t\to 0$$$$\underset{t\to 0}{\lim }{\left(\frac{\sin (t+3)}{\sin 3}\right)}^{\frac{1}{t}}=\underset{t\to 0}{\lim }{(1+\left(\frac{\sin (t+3)-\sin 3}{\sin 3}\right))}^{\frac{1}{t}}$$$$=e^{\underset{t\to 0}{\lim }\left(\frac{\sin (t+3)-\sin 3}{t.\sin 3}\right)}=$$$$e^{\underset{t\to 0}{\lim }\left(\frac{2\sin \left(\frac{t}{2}\right)\cos (\frac{t}{2}+3)}{t.\sin 3}\right)}=e^{\cot 3}.theans$$

Commented by jagoll last updated on 09/Feb/20

$$wawthankyousir$$

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