Find without L'Hospital's rule: $\lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}}$ $A = \sqrt{17 - 2x^{2}} \sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} \\ 
\Rightarrow \lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}} \\ 
=  \lim\limits_{x \to 2} \frac{A^{6} - 9^{6}}{(x - 2)^{2}(A^{5} + 9A^{4} + ... + 9^{5})} \\ 
=  \lim\limits_{x \to 2} \frac{(x - 2)^{2}(-72x^{10} - 192x^{9} + 940x^{8} + 2576x^{7} + 874x^{6} + 1992x^{5} - 24543x^{4} - 73908x^{3} - 82540x^{2} - 200414x - 102154)}{(x - 2)^{2}(A^{5} + 9A^{4} + ... + 9^{5})} \\ 
=  \frac{-2.11.3^{10}}{6.9^{5}} = \frac{-11}{3}.$
Am I right? Is there a simple way?

Yes, that works.
You can also use the fact that, if $f(x)=\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}$, then $f(2)=9$, $f'(2)=0$, and $f''(2)=-\frac{22}3$. So, near $0$,$$f(x)=9-\frac{11}3(x-2)^2+O\bigl((x-2)^3\bigr)$$ and therefore,\begin{align}\lim_{x\to2}\frac{\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}}{(x-2)^2}&=\lim_{x\to2}\frac{f(x)-f(2)}{(x-2)^2}\\&=-\frac{11}3.\end{align}
