Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $

If you are allowed to use the binomial expansion, set $y=x-7$ and examine to first order$$\frac {3\sqrt {1+\frac y9}-3\sqrt[3]{1+\frac y{27}}}{2\sqrt[4]{1+\frac y{16}}-2}=\frac {\frac 32\cdot\frac y9-\frac 33\cdot\frac y{27}}{\frac 24\cdot\frac y{16}}=\frac {112}{27}$$

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.

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

$$32\lim_{x\to 7}\frac{(x+2)^3-(x+20)^2}{x-7}\frac{1}{(3+3)(3^2-3\cdot 3+3^2)(3^2+3\cdot 3+3^2)}$$

$$=\frac{32}{1458}\lim_{x\to 7}\frac{x^3+5x^2-28x-392}{x-7}=\frac{32}{1458}\lim_{x\to 7}x^2+12x+56=\frac{32}{1458}(189)=\frac{112}{27}$$