Calculate $\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$ Please help me calculate this:
$$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$
Here I've tried multiplying by $\sqrt[4]{x+9}+2$ and few other method. 
Thanks in advance for solution / hints using simple methods.
Edit
Please don't use l'Hosplital rule. We are before derivatives, don't know how to use it correctly yet. Thanks!

Hint: Use that $\frac{a^4-b^4}{a-b} = a^3 + a^2b + ab^2 + b^3$. Setting $a=\sqrt[4]{x+9}$ and $b=2$, you see $a^4-b^4 = x-7$, and you get $$\frac{1}{\sqrt[4]{x+9}-2} = \frac{a^3 + a^2b + ab^2 + b^3}{x-7}$$
Similarly you can write $\sqrt{x+2}-3$  as:
$$\sqrt{x+2}-3 = \frac{x-7}{\sqrt{x+2}+3}$$
And a similar but uglier result for $\sqrt[3]{x+20}-3$ using $u-v=\frac{u^3-v^3}{u^2+uv+v^2}$ with $u=\sqrt[3]{x+20}$ and $v=3$, that gives $u^3-v^3 = x-7$.
Note then that the $x-7$s cancel out, and you get an expression where none of the numerators or denominators approach zero as $x\to 7$, so you can finally just plug in $x=7$ in that expression.
Cancelling out the $x-7$ terms, you get:
$$(a^3 + a^2b + ab^2 + b^3)\left(\frac{1}{\sqrt{x+2}+3}-\frac{1}{u^2+uv+v^2}\right)$$
But as $x\to 7$ $a\to b=2$ and $u\to v=3$.
So the limit is:
$$(4\cdot2^3)\left(\frac{1}{6}-\frac{1}{3\cdot 3^2}\right)$$
