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.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
 A: If you are allowed to use the binomial expansion, set $y=x-7$ and examine to first order$$\frac {3\sqrt {1+\frac y9}-3\sqrt[3]{1+\frac y{27}}}{2\sqrt[4]{1+\frac y{16}}-2}=\frac {\frac 32\cdot\frac y9-\frac 33\cdot\frac y{27}}{\frac 24\cdot\frac y{16}}=\frac {112}{27}$$
A: 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. 
A: 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
$$32\lim_{x\to 7}\frac{(x+2)^3-(x+20)^2}{x-7}\frac{1}{(3+3)(3^2-3\cdot 3+3^2)(3^2+3\cdot 3+3^2)}$$
$$=\frac{32}{1458}\lim_{x\to 7}\frac{x^3+5x^2-28x-392}{x-7}=\frac{32}{1458}\lim_{x\to 7}x^2+12x+56=\frac{32}{1458}(189)=\frac{112}{27}$$
A: Sorry to add a second answer, dont upvote if you dont like it, but this answer is different from my first and is essentially the same as some of the other answers, but I wanted to point a different formulation that is a little to long for a comment.
One can use the limit
$$\lim\limits_{y \to 0} \frac{(y+k)^{\alpha}-k^{\alpha}}{y}=\alpha k^{\alpha -1}$$
then with the substitution, $y=x-7$ we can write the expression as
$$\frac{\left( \frac{\sqrt{y+9}-3}{y} \right) -\left( \frac{\sqrt[3]{y+27}-3}{y} \right)}{\frac{\sqrt[4]{y+16}-2}{y} }$$
and this gives, 
$$\frac{\frac{1}{2}9^{-\frac{1}{2}}-\frac{1}{3}27^{-\frac{2}{3}} }{\frac{1}{4}16^{-\frac{3}{4}}}=\frac{112}{27}$$
A: Letting $x = 7 + h$, then we find: $$\displaystyle \lim_{h \to 0} \dfrac{\sqrt{h+9} - \sqrt[3]{h+27}}{\sqrt[4]{h+16} - 2} = \displaystyle \lim_{h \to 0} \dfrac{3\left(1+\dfrac{h}{18}\right) - 3\left(1+\dfrac{h}{81}\right)}{2\left(1+\dfrac{h}{64}\right) - 2} = \dfrac{112}{27},$$
where $(1+h)^{\alpha} \sim_0 1 + \alpha h$.
