I found difficulties by finding a limit $$\lim_{x\to 3} \left(\frac{\sin(x)}{\sin(3)}\right)^{\frac{1}{x-3}} $$ without using L'Hospital's role or any of derivative definition at all.$$$$ I was trying to use formulas $\lim_{x\to 0} \left(\frac{\sin(x)}{x}\right) $ and $ \sin(x)=\sin(x-3+3)=\sin(x-3)\cos(3)+\cos(x-3)\sin(3) $ but nothing good happened. $$ $$Later on I was wondering about $\lim_{x\to a}{f(x)^{g(x)}}=e^{\lim_{x\to a}{g(x) \cdot ln(f(x)})}$ but also have no idea... Can anyone help?
Finding limit $\lim_{x\to 3} \left(\frac{\sin(x)}{\sin(3)}\right)^{\frac{1}{x-3}} $ without using L'Hospital role
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On
First rewrite the expression as $e^{\frac{\log (\sin x) - \log \sin 3}{x-3}}$ and then set $t=x-3$ to get in the argument of $e(\cdot )$: $$ f(3,t) = \frac{\log \sin (3+t) - \log \sin 3}{t} $$ and note that $e(\cdot)$ is a continuous function, so you can interchange $\exp$ and the limit. The expression above is a definition of the derivative: $f'(3)$. So take this derivative and get the final answer: $ e^{\frac{\cos 3}{\sin 3}}$.
EDIT: to make things a bit more precise, here, what we have is a limit: $$ \lim_{y \to 3} f (y,x) = \lim_{y \to 3} \lim_{t \to 0} \frac{\log \sin (y+t) - \log \sin y}{t} $$ So this is a definition of the derivative, i.e. a limit. Once you take it, you get the result.
On
$$ \ln(L) =\lim _{x\to 3}\left(\frac{\ln \left(\frac{\sin \left(x\right)}{\sin \left(3\right)}\right)}{x-3}\right) = \lim _{y\to 0}\left(\frac{\ln \left(\frac{\sin \left(y+3\right)}{\sin \left(3\right)}\right)}{y}\right) \color{red}{=_H} \lim _{y\to 0}\left(\cot \left(y+3\right)\right) = \cot \left(3\right) $$ So
$$L = e^{\cot(3)}$$ where $\color{red}{=_H}$ is L'Hôpital's rule
$$\left[\lim_{\sin x\to\sin3}\left(1+\dfrac{\sin x-\sin3}{\sin 3}\right)^{1/(\sin x-\sin3)}\right]^{\lim_{x-3}\frac{\sin x-\sin3}{x-3}}$$
The inner limit converges to $e$
For the exponent, $$\lim_{y-a}\frac{\sin2y-\sin2a}{2y-2a}=\lim_{y-a}\dfrac{\sin(y-a)}{y-a}\cdot\lim_{y-a}\sin(y+a)=?$$
Here $2y=x,2a=3$