<s>[INST] How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:

$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$

Is there a way to calculate it? How can I do it?
 [/INST] With Taylor's formula at order $1$:
Set $x=7+h$. You obtain
\begin{align}
\frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}&=\frac{\sqrt{9+h} - \sqrt[3]{27+h}}{\sqrt[4]{16+h} - 2}=\frac{3\sqrt{1+\frac h9} - 3\sqrt[3]{1+\frac h{27}}}{2\sqrt[4]{1+\frac h{16}} - 2}\\
&=\frac{\bigl(3+\frac h6+o(h)\bigr)-\bigl(3+\frac h{27}+o(h)\bigr)}{2+\frac h{32}+o(h)-2}=\frac{\frac{7h}{54}+o(h)}{\frac h{32}+o(h)}=\frac{\frac{7}{54}+o(1)}{\frac 1{32}+o(1)}\to \frac{112}{27}.
\end{align}
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