<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] The unpopular limit formula $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1}\tag{1}$$ is your friend here. First divide the numerator and denominator of the expression by $x-7$ and note that the denominator itself becomes a fraction $$\frac{\sqrt[4]{x+9}-2} {x-7}=\frac{t^{1/4}-16^{1/4}}{t-16}$$ where $t=x+9\to 16$. By the limit formula $(1)$ the above fraction tends to $(1/4)16^{-3/4}=1/32$.
Next consider the numerator which has now become $$\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}=\frac{u^{1/2}-9^{1/2}}{u-9}-\frac{v^{1/3}-27^{1/3}}{v-27}$$ where $u=x+2\to 9,v=x+20\to 27$. By the limit formula $(1)$ the above expression tends to $(1/2)9^{-1/2}-(1/3)27^{-2/3}=7/54$. The final answer is now $(7/54)/(1/32)=112/27$.
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