<s>[INST] How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
 [/INST] $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\lim_{x\to 1} \frac{\sqrt{3+x}-2}{x-1}\cdot \lim_{x\to 1} \frac{x-1}{\sqrt[3]{7+x}-2}$$ $$=\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}$$
Use $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$  which gives  
$$\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}=(0.5)4^{-0.5}\cdot\frac{1}{(1/3)8^{-2/3}}=3$$
see 
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