$\displaystyle
\displaystyle\lim_{x\to\infty}\sqrt[3]{(x^3+5x^2)}-x
$
If you know these methods you can try:
$\displaystyle
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x
$
Expand the cube root using the binomial theorem:
$\displaystyle
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x=x~\left(1+\frac{5}{3x}+O(x^{-2})\right) -x=\frac{5}{3}+O(x^{-1})
$
Hence the required limit is $\displaystyle 5/3$.
RonL
Using your method gives me this:
$\displaystyle
\lim_{x\to\infty}\dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{\frac{1}{x^3} + \frac{5}{x^4} }} + 1 }
$
and when you take the limit to infinity you get $\displaystyle \frac{5}{2}$ but the answer is $\displaystyle \frac{5}{3}$
Check that again:
$\displaystyle \lim_{x \to \infty} \dfrac{5x^2}{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }$
$\displaystyle = \lim_{x \to \infty} \dfrac{5}{\frac{1}{x^2}\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + \frac{1}{x}\sqrt[3]{{x^3 + 5x^2 }} + 1 }$
$\displaystyle = \lim_{x \to \infty} \dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{1 + \frac{5}{x} }} + 1 }$
$\displaystyle = \frac{5}{1 + 1 + 1} = \frac{5}{3}$
-Dan