Please help me calculate $$\lim_{n\to \infty}{\frac{\sqrt[3]{n^4+2}-\sqrt[3]{n^4+4n^3-1}}{\sqrt[3]{n+4}}} $$
I don’t understand very well what needs to be done here with roots. I tried to multiply the entire expression by the value of the numerator and denominator and divide the entire expression by $n^\frac{4}{3}$, but the resulting expressions were very large and I still did not understand what to do with them...
I don't know what L'Hôpital's rule is. We can't use it yet. Please don't use it if you try to explain this to me.
Hint
$$\frac{\sqrt[3]{n^4+2}-\sqrt[3]{n^4+4n^3-1}}{\sqrt[3]{n+4}}=\frac{n^{4/3}\, \sqrt[3]{1+\frac 2 {n^4} }-n^{4/3}\, \sqrt[3]{1+\left(\frac 4 n -\frac 1 {n^4}\right) }} {n^{1/3}\, \sqrt[3] {{1+\frac 4 {n} } }}$$ So, you face three times $$\large\sqrt[3] {1+\epsilon }$$
Use the binomial expansion and simplify as much as you can.