<s>[INST] Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle  \lim_{x \to 3}  \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).
 [/INST] $$\frac{\sqrt{19-x}-2 \sqrt[4]{x+13}}{\sqrt[3]{11-x}-x+1}$$
plug $x=y+3$ so we get
$$\underset{y\to 0}{\text{lim}}\frac{\sqrt{16-y}-2 \sqrt[4]{y+16}}{\sqrt[3]{8-y}-y-2}$$
we have the following expansions as $y\to 0$
$$
\begin{array}{rll}
 \sqrt{16-y} &\sim &4-\frac{y}{8} \\
 2 \sqrt[4]{y+16} &\sim& \frac{y}{16}+4 \\
 \sqrt[3]{8-y} &\sim& 2-\frac{y}{12} \\
\end{array}
$$
So the limit can be written as
$$\underset{y\to 0}{\text{lim}}\frac{4-\frac{y}{8}-\left(\frac{y}{16}+4\right)}{2-\frac{y}{12}-y-2}=\lim_{y\to0}\frac{-\frac{3}{16}y}{-\frac{13}{12}y}=\frac{9}{52}$$
</s>