<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] For convenience, change variables to $t = x - 1$, so you're looking at 
$\dfrac{\sqrt{4+t}-2}{(8+t)^{1/3}-2}$.
$$ \sqrt{4+t} = 2 \sqrt{1+\frac{t}{2}} = 2 \left(1 + \frac{t}{4} + \ldots\right) $$
$$ (8+t)^{1/3} = 2 \left(1 + \frac{t}{8}\right)^{1/3} = 2 \left(1 + \frac{t}{24} + \ldots\right) $$
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