<s>[INST] Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
 [/INST] Here is another solution, not using Taylor series, but based on multiplying by conjugates.
I wanted to see if it would work.
Note $x^3-y^3=(x-y)(x^2+xy+y^2)$
so $$\sqrt{x+2}-\sqrt[3]{x+20}=\frac{\sqrt{(x+2)^3}-(x+20)}{x+2+ \sqrt{x+2}\sqrt[3]{x+20}+\sqrt[3]{(x+20)^2}}
$$and
$$\sqrt{(x+2)^3}-(x+20)=\frac{(x+2)^3-(x+20)^2}{\sqrt{(x+2)^3}+(x+20)}$$
and 
$$(x+2)^3-(x+20)^2=(x-7)(x^2+12x+56)$$
For the denominator we have 
$$\sqrt[4]{x+9}-2=\frac{\sqrt{x+9}-4}{\sqrt[4]{x+9}+2}
=\frac{x-7}{(\sqrt[4]{x+9}+2)(\sqrt{x+9}+4)}
$$
So all together,
$$\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
=\frac{(x^2+12x+56)(\sqrt[4]{x+9}+2)(\sqrt{x+9}+4) }{(x+2+ \sqrt{x+2}\sqrt[3]{x+20}+\sqrt[3]{(x+20)^2})(\sqrt{(x+2)^3}+(x+20))}$$
Now if you substitute $x=7$ you get $\frac{112}{27}$ like the other answers. 
</s>