<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] Here is a simple strategy. Find the limits of those radicals and subtract that limit from them.
Thus for numerator the term $\sqrt{19-x}\to 4$ and hence subtract and add $4$ to get $$\sqrt{16-x}-4+4$$ The other term in numerator is $2\sqrt[4]{13+x}$ which tends to $4$ and replace it with $$2\sqrt[4]{13+x}-4+4$$ and see that that those extra $4$'s added in both terms cancel out (well such limit problems are designed in such manner that numerator and denominator tend to $0$ and thus we have the desired cancelation here). Thus numerator equals $$\sqrt{19-x}-4-(2\sqrt[4]{13+x}-4)$$ and similarly denominator equals $$\sqrt[3]{11-x}-2-(x-3)$$ Now put $x-3=h$ so that $h\to 0$ and divide each term in numerator and denominator by $h$ to get $$\lim_{h\to 0}\dfrac{\dfrac{\sqrt{16-h}-4}{h}-2\cdot\dfrac{\sqrt[4]{16+h}-2}{h}}{\dfrac{\sqrt[3]{8-h}-2}{h}-1}$$ You can now evaluate the limit of three fractions easily and get the final answer as $$\frac{(-1/8)-2(1/32)} {(-1/12)-1} =\frac{9}{52}$$
</s>