# Solve. lim_(x rarr oo)(root(3)(x^3-2x^2+3)/(2x+1)

Solve. $\underset{x\to \mathrm{\infty }}{lim}\left(\frac{\sqrt[3]{{x}^{3}-2{x}^{2}+3}}{2x+1}$
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faldduE
$\underset{x\to \mathrm{\infty }}{lim}\left(\left(\frac{\sqrt[3]{{x}^{3}-2{x}^{2}+3}}{2x+1}\right)=\frac{1}{2}$
To start, divide by the highest denominator power by using the following algebraic property:
$a+b=a\left(1+\frac{b}{a}\right)$
$\left(\sqrt[3]{{x}^{3}-2{x}^{2}+3}=\left(\sqrt[3]{{x}^{3}\left(1-\frac{2}{x}+\frac{3}{{x}^{3}}\right)}$
$x\to \mathrm{\infty }⇒\sqrt[3]{{x}^{3}}=x$
$x\sqrt[3]{1-\frac{2}{x}+\frac{3}{{x}^{3}}}$
$=\frac{\sqrt[3]{1-\frac{2}{x}+\frac{3}{{x}^{3}}}}{2x+1}$
Then, divide by x and refine:
$\frac{\sqrt[3]{1-\frac{2}{x}+\frac{3}{{x}^{3}}}}{x}/\left(2\frac{x}{2}+\frac{1}{x}\right)$
$=\frac{\sqrt[3]{1-\frac{2}{x}+\frac{3}{{x}^{3}}}}{2+\frac{1}{x}}$
Then, take the limit of the numerator and denominator using the limit quotient rule, which is:
$\underset{x\to \mathrm{\infty }}{lim}\left[\left(\frac{f\left(x\right)}{g\left(x\right)}\right)\right]=\frac{\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)}{\underset{x\to \mathrm{\infty }}{lim}g\left(x\right)},\underset{x\to \mathrm{\infty }}{lim}g\left(x\right)$
$\frac{lim\left(x\to \mathrm{\infty }\right)\left(\sqrt[3]{1-\frac{2}{x}+\frac{3}{{x}^{3}}}\right)}{\underset{x\to \mathrm{\infty }}{lim}\left(2+\frac{1}{x}\right)}$
The limit of the numerator is 1:
$\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt[3]{1-\frac{2}{x}+\frac{3}{{x}^{3}}}\right)$
$\sqrt[3]{\underset{x\to \mathrm{\infty }}{lim}\left(1-\frac{2}{x}+\frac{3}{{x}^{3}}\right)}$
$\sqrt[3]{\underset{x\to \mathrm{\infty }}{lim}\left(1-\underset{x\to \mathrm{\infty }}{lim}\left(\frac{2}{x}+\underset{x\to \mathrm{\infty }}{lim}\left(\frac{3}{{x}^{3}}\right)\right)\right)}$