How to evaluate $\lim_{n->\infty}\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{3}{n}}}{\sqrt[6]{n^2+\sin{\frac{2}{n}}}-\sqrt[3]{n}}?$ 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?
$\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}=$
$=\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}\dfrac{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}$
$=\lim_{n->\infty}\dfrac{1-\cos\dfrac{3}{n}}{(\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n})((\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2)}$
 A: 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
$$\frac{6}{3} \cdot \frac{9}{4} = \frac{9}{2}.$$
A: As said in comments, using Taylor series would make life much easier.
Start dividing all terms by $n^{\frac 13}$ and let $x=\frac 1n$. Now, use the well known Taylor series for the sine and the cosine. Continue with Taylor or with the binomial expansion.
Using three terms in the expansions, you should end with the limit itself. 
Adding an extra term in the expansion, you will end with
$$\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{3}{n}}}{\sqrt[6]{n^2+\sin{\frac{2}{n}}}-\sqrt[3]{n}}=\frac{9}{2}-\frac{3}{n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
If you make $n=10$ (quite far away from $\infty$), using you pocket calculator (I made it on my phone), you should get an exact value of $\approx 4.22868$ while the above truncated expression would give $\frac{21}{5}=4.2$.
Making the problem more general as  in a now deleted answer, the same process would give
$$\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{a}{n}}}{\sqrt[6]{n^2+\sin{\frac{b}{n}}}-\sqrt[3]{n}}=\frac{a^2}{b}-\frac{2 a^2}{3 b n}+O\left(\frac{1}{n^2}\right)$$
