Question

1/ \(\lim\limits_{x\to 1}\) \(\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}\)

2/ \(\lim\limits_{x \to \ +\infty} \)\(x\left[\sqrt{4x^2+5}-\sqrt[3]{8x^3-1}\right]\)

3/ \(\lim\limits_{x\to 1}\)\(\dfrac{x^3-2x-1}{x^5-2x-1}\)

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Answer 1

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).