How can I calculate this limit: $\lim\limits_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$? I was given this limit to solve, without using L'Hospital rule. It's killing me  !! Can I have the solution please ?
$$\lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$$
 A: $$ \lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} $$
$$ = \lim_{x\rightarrow 2} \frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} \times \frac{2+\sqrt{2+x}}{2+\sqrt{2+x}} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3} }{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}$$
$$ = \lim_{x\rightarrow 2} \frac{2 - x}{-(2-x)} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}{2 + \sqrt{2+x}}$$
$$= -\frac{3 (2) ^{2/3}}{4} = -\frac{3}{2(2)^{1/3}}$$
A: The solution below may come perilously close to the forbidden L'Hospital's Rule, though the Marquis is not mentioned.
To make things look more familiar, change signs, and then divide top and bottom by $x-2$. The expression we want the limit of becomes
$$\frac{\sqrt{2+x}-2}{x-2} \cdot \frac{x-2}{(4-x)^{1/3}-2^{1/3}}.$$
We recognize the first part of the product as the "difference quotient" $\frac{f(x+a)-f(a)}{x-a}$ where $f(x)=\sqrt{2+x}$ and $a=2$.
We recognize the second part of the product as the reciprocal of the difference quotient $\frac{g(x+a)-g(a)}{x-a}$ where $g(x)=(4-x)^{1/3}$ and $a=2$.
Now take the derivatives. 
