<s>[INST] 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 ?
$$\lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$$
 [/INST] $$ \lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} $$
$$ = \lim_{x\rightarrow 2} \frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} \times \frac{2+\sqrt{2+x}}{2+\sqrt{2+x}} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3} }{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}$$
$$ = \lim_{x\rightarrow 2} \frac{2 - x}{-(2-x)} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}{2 + \sqrt{2+x}}$$
$$= -\frac{3 (2) ^{2/3}}{4} = -\frac{3}{2(2)^{1/3}}$$
</s>