<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] The conjugation/rationalization approach works, but is somewhat tedious. The key is to introduce the right terms in order to force the numerator and denominator into a differences of integer powers:
$$\begin{align}
\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}&=\frac{\left((19-x)^2\right)^{\frac14}-\left(2^4(13+x)\right)^{\frac14}}{(11-x)^{\frac13}-\left((x-1)^3\right)^{\frac13}}\\[1ex]
&=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\\[1ex]
&=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\times\frac{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}\\[1ex]
&=\frac{\left(a^{\frac14}\right)^4-\left(b^\frac14\right)^4}{\left(c^{\frac13}\right)^3-\left(d^\frac13\right)^3}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\\[1ex]
&=\frac{a-b}{c-d}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}
\end{align}$$
(where I hope the replacements of $a,b,c,d$ are obvious)
Upon simplification (with $x\neq3)$, we have
$$\frac{a-b}{c-d}=\frac{153-54x+x^2}{12-4x+3x^2-x^3}=\frac{(51-x)(3-x)}{(3-x)(4+x^2)}=\frac{51-x}{4+x^2}$$
and the remaining fraction of rational powers is continuous at $x=3$. Then the limit is
$$\lim_{x\to3}\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}=\frac{48}{13}\times\frac{12}{256}=\boxed{\frac9{52}}$$
</s>