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?

Using $x^n - y^n = (x - y)\sum_{k=0}^n x^ky^{n-k}$, we have
$$
\sqrt{x+2}-\sqrt[3]{x+20} = \frac{(x+2)^3-(x+20)^2}{\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}} \\
\sqrt[4]{x+9}-2 = \frac{x-7}{\sum_{k=0}^4(x+9)^{k/4}2^{4-k}}
$$
Notice that $(x+2)^3-(x+20)^2 = (x - 7)(x^2+12x+56)$, so
$$
\begin{aligned}
\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} & = \frac{(x - 7)(x^2+12x+56)/\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}}{(x-7)/\sum_{k=0}^4(x+9)^{k/4}2^{4-k}} \\
& = (x^2+12x+56)\frac{\sum_{k=0}^4(x+9)^{k/4}2^{4-k}}{\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}}
\end{aligned}
$$
so
$$
\begin{aligned}
\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} & = (7^2+12\cdot7+56)\frac{\sum_{k=0}^416^{k/4}2^{4-k}}{\sum_{k=0}^69^{k/2}27^{2-k/3}} \\
& = 189\cdot\frac{2^4}{27^2} = \frac{112}{27}
\end{aligned}
$$
