Calculate $\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin\left(x-3\right)}}$ without L'Hôspital I have to calculate this limit $$\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin (x-3)}}$$
without L'Hôpital's rule.
These are my steps
$$\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\:=\:e^{\ln\left(\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\right)}=e\:^{\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}}$$
now i will caculate only $\frac{x\left(3x-8\right)}{\sin\left(x-3\right)}$
$$ t = x-3 \Rightarrow x = t+3$$
$\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\:=\lim \:_{t\to \:0}\:\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}$
And here im stack.
Help someone ?
Thanks.
 A: Note that
\begin{align}
\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\ &= \lim_{t\to 0}\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}\\
&= \lim_{t\to 0}\frac{\left(t+3\right)\ln\left(3t+9-8\right)}{\sin\left(t\right)}\\
&= \lim_{t\to 0}\frac{\left(3\right)\ln\left(1 + 3t\right)}{\sin\left(t\right)} \cdot \frac{t}{t} \cdot \frac{3t}{3t}\\
&=  \lim_{t\to 0} 3\frac{3t}{t}\\
&= 9
\end{align}
thus
$$\lim_{x\to 3}\left(3x-8\right)^{\frac{x}{\sin (x-3)}} = \lim_{x \to 3} e^{\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}} = e^9$$
A: Using the Taylor expansion note that $$
\begin{align}
\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\ &= \lim_{t\to 0}\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}\\
&= \lim_{t\to 0}\frac{(t+3)(3t-\frac{3t^2}{2}+\frac{3t^3}{3!}+o({t^4)}}{t-\frac{t^3}{3!}+o(t^4)}
=9\end{align}$$
A: Rewrite
$$
\frac{x \ln(3x-8)}{\sin(x-3)} = x \frac{\ln(1+(3x-9))}{3x-9}\frac{x-3} {\sin(x-3)}\frac{3x-9}{x-3}\to3\cdot1\cdot1\cdot 3=9
$$
as $x\to3$. 
A: You can go in a more direct manner since $3x-8$ is bounded when $x\to 3$.
$$\lim\limits_{x\to 3}(3x-8)^{\frac{x}{\sin(x-3)}}=\lim\limits_{x-3=t\to 0}(9+3t-8)^{\frac{t+3}{\sin(t)}}=\lim\limits_{t\to 0}(1+3t)^{\frac{t}{\sin(t)}+\frac{3}{\sin(t)}}=\lim\limits_{t\to 0}(1+3t)^{1+\frac{3}{\sin(t)}}=\lim\limits_{t\to 0}(1+3t)^{1}(1+3t)^{\frac{3}{\sin(t)}}=\lim\limits_{t\to 0}(1+3t)^{\frac{3}{\sin(t)}\frac{t}{t}}=\lim\limits_{t\to 0}(1+3t)^{\frac{t}{\sin(t)}\frac{3}{t}}=\lim\limits_{t\to 0}(1+3t)^{\frac{3}{t}}=\lim\limits_{3t=u\to 0}(1+u)^{\frac{3}{\frac{u}{3}}}=\lim\limits_{u\to 0}(1+u)^{\frac{9}{u}}=e^9$$
