I tried to get rid off cube root as written below but still can not get throught the next steps. What should be the right step to take after the steps below? Did I start as I should or do I have to take completely different approach?
$\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}=$
$=\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}\dfrac{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}$
$=\lim_{n->\infty}\dfrac{1-\cos\dfrac{3}{n}}{(\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n})((\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2)}$
Best Answer
To complete your approach, let \begin{align*} a &= \frac{\sqrt[3]{n + 1}}{\sqrt[3]{n}} = \sqrt[3]{1 + \frac{1}{n}} \\ b &= \frac{\sqrt[3]{n + \cos \frac{3}{n}}}{\sqrt[3]{n}} = \sqrt[3]{1 + \frac{1}{n}\cos\frac{3}{n}} \\ c &= \frac{\sqrt[6]{n^2 + \sin \frac{2}{n}}}{\sqrt[3]{n}} = \sqrt[6]{1 + \frac{1}{n^2}\sin\frac{2}{n}} \\ d &= \frac{\sqrt[3]{n}}{\sqrt[3]{n}} = 1 \end{align*}
Then, your expression becomes, $$\frac{a - b}{c - d} = \frac{(a^3 - b^3)(c^5 + c^4d + c^3d^2 + c^2 d^3 + cd^4 + d^5)}{(c^6 - d^6)(a^2 + ab + b^2)}.$$
Note that $a, b, c, d \to 1$ as $n \to \infty$, so $$\lim \frac{a - b}{c - d} = \frac{6}{3} \lim \frac{a^3 - b^3}{c^6 - d^6},$$ assuming the limit on the right exists. We have, $$\frac{a^3 - b^3}{c^6 - d^6} = \frac{\frac{1}{n}\left(1 - \cos \frac{3}{n}\right)}{\frac{1}{n^2} \sin\frac{2}{n}} = \frac{\frac{1}{n}\left(1 - \cos^2 \frac{3}{n}\right)}{\frac{1}{n^2} \sin\frac{2}{n}\left(1 + \cos \frac{3}{n}\right)} = \frac{\frac{\sin^2 \frac{3}{n}}{\frac{9}{n^2}} \cdot 9}{\frac{\sin \frac{2}{n}}{\frac{2}{n}} \cdot 2 \cdot(1 + \cos \frac{3}{n})} \to \frac{9}{4}.$$ Hence, our complete limit is $$\frac{6}{3} \cdot \frac{9}{4} = \frac{9}{2}.$$