Solving Limit Problem with N Approaching Infinity

[Mutlu]

Hello,
Here is my example, could you please check and correct if it needed.
$\stackrel{lim}{n\rightarrow ∞}\frac{\sqrt{4n^3+1}-\sqrt[3] {n+2}}{\sqrt[3] {n^6+27}+n}=$ $\stackrel{lim}{n\rightarrow ∞}\frac{\sqrt{4n^3}}{n\sqrt[3] {n^6}}=$$\stackrel{lim}{n\rightarrow ∞}{\frac{{{4n}^\frac{3}{2}}}{n^2}}={{4}^\frac{3}{4}}$
Thank you!

 

[HallsofIvy]

I don't know how to correct that. Your last two steps, from the limit of $\sqrt{4n^3}/n\sqrt[3]{n^6}$ to the limit of $4n^{3/2}/n^2$ and then to $4^{3/2}$ are pretty much nonsense. Do them again!

 

[Mutlu]

I don't know how to correct that. Your last two steps, from the limit of $\sqrt{4n^3}/n\sqrt[3]{n^6}$ to the limit of $4n^{3/2}/n^2$ and then to $4^{3/2}$ are pretty much nonsense. Do them again!

$\stackrel{lim}{n\rightarrow ∞}\frac{\sqrt{4n^3}}{n\sqrt[3] {n^6}}=$$\stackrel{lim}{n\rightarrow ∞}{\frac{{{n}^\frac{3}{2}}}{n^2}}=\stackrel{lim}{n\rightarrow ∞}{\frac{{1}}{{n}^\frac{1}{2}}}=0$
Like this?

 

[dextercioby]

Now it looks ok.

 

[Mutlu]

A little a bit late, but anyway Thanks a lot!