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).
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Let $y=x-3$.
Then it's $\lim\limits_{y\to0}\dfrac{\sqrt{16-y}-2\sqrt[4]{16+y}}{\sqrt[3]{8-y}-y-2}.$
Using binomial series, that's $\dfrac{(4-\frac18y\dots)-2(2+\frac1{32}y\dots)}{(2-\frac1{12}y...)-y-2}=\dfrac{-3/16}{-13/12}=\dfrac9{52.}$
On
HINT.-Use the identity $A^3-B^3=(A-B)(A^2+AB+B^2)$ with $A=\sqrt[3]{11-x}$ and $B=x-1$ and multiplying numerator and denominator by $(A^2+AB+B^2)$ you' ll get again the undeterminated form $\dfrac{(4-4)(4+4+4)}{8-8}$. But the new denominator being equal to $11-x-(x-1)^3$ you have its derivative is equal to $-13\ne0$ so the undetermination disappear after apply Hospital's rule.
On
$$\frac{\sqrt{19-x}-2 \sqrt[4]{x+13}}{\sqrt[3]{11-x}-x+1}$$ 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 $$\underset{y\to 0}{\text{lim}}\frac{4-\frac{y}{8}-\left(\frac{y}{16}+4\right)}{2-\frac{y}{12}-y-2}=\lim_{y\to0}\frac{-\frac{3}{16}y}{-\frac{13}{12}y}=\frac{9}{52}$$
On
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 $$\frac{(-1/8)-2(1/32)} {(-1/12)-1} =\frac{9}{52}$$
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
$$\lim_{x\to3}\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}=\frac{48}{13}\times\frac{12}{256}=\boxed{\frac9{52}}$$