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.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$

Because $(a-b)(a+b)(a^2+b^2)=a^4-b^4$, and $\lim_{x\to 7}a=\lim_{x\to 7}b=2$, we rewrite as
$$\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}=\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{(x+9)-16}(2+2)(2^2+2^2)$$
$$=32\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}$$
Because $(a-b)(a+b)(a^2+ab+b^2)(a^2-ab+b^2)=a^6-b^6$, and also $\lim_{x\to 7}a=\lim_{x\to 7}b=3$, we rewrite as
$$32\lim_{x\to 7}\frac{(x+2)^3-(x+20)^2}{x-7}\frac{1}{(3+3)(3^2-3\cdot 3+3^2)(3^2+3\cdot 3+3^2)}$$
$$=\frac{32}{1458}\lim_{x\to 7}\frac{x^3+5x^2-28x-392}{x-7}=\frac{32}{1458}\lim_{x\to 7}x^2+12x+56=\frac{32}{1458}(189)=\frac{112}{27}$$
