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

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

Question
Solve. \(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left(\frac{{\sqrt[{{3}}]{{{x}^{{3}}-{2}{x}^{{2}}+{3}}}}}{{{2}{x}+{1}}}\right.}\)

Answers (1)

2021-02-26
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({\left(\frac{{\sqrt[{{3}}]{{{x}^{{3}}-{2}{x}^{{2}}+{3}}}}}{{{2}{x}+{1}}}\right)}=\frac{{1}}{{2}}\right.}\)
To start, divide by the highest denominator power by using the following algebraic property:
\(\displaystyle{a}+{b}={a}{\left({1}+\frac{{b}}{{a}}\right)}\)
\(\displaystyle{\left({\sqrt[{{3}}]{{{x}^{{3}}-{2}{x}^{{2}}+{3}}}}={\left({\sqrt[{{3}}]{{{x}^{{3}}{\left({1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}\right)}}}}\right.}\right.}\)
\(\displaystyle{x}\rightarrow\infty\Rightarrow{\sqrt[{{3}}]{{{x}^{{3}}}}}={x}\)
\(\displaystyle{x}{\sqrt[{{3}}]{{{1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}}}}\)
\(\displaystyle=\frac{{{\sqrt[{{3}}]{{{1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}}}}}}{{{2}{x}+{1}}}\)
Then, divide by x and refine:
\(\displaystyle\frac{{{\sqrt[{{3}}]{{{1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}}}}}}{{x}}/{\left({2}\frac{{x}}{{2}}+\frac{{1}}{{x}}\right)}\)
\(\displaystyle=\frac{{{\sqrt[{{3}}]{{{1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}}}}}}{{{2}+\frac{{1}}{{x}}}}\)
Then, take the limit of the numerator and denominator using the limit quotient rule, which is:
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left[{\left(\frac{{f{{\left({x}\right)}}}}{{g{{\left({x}\right)}}}}\right)}\right]}=\frac{{\lim_{{{x}\rightarrow\infty}}{f{{\left({x}\right)}}}}}{{\lim_{{{x}\rightarrow\infty}}{g{{\left({x}\right)}}}}},\lim_{{{x}\rightarrow\infty}}{g{{\left({x}\right)}}}\)
\(\displaystyle\frac{{\lim{\left({x}\rightarrow\infty\right)}{\left({\sqrt[{{3}}]{{{1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}}}}\right)}}}{{\lim_{{{x}\rightarrow\infty}}{\left({2}+\frac{{1}}{{x}}\right)}}}\)
The limit of the numerator is 1:
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({\sqrt[{{3}}]{{{1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}}}}\right)}\)
\(\displaystyle{\sqrt[{{3}}]{{\lim_{{{x}\rightarrow\infty}}{\left({1}-\frac{{2}}{{x}}+\frac{{3}}{{x}^{{3}}}\right)}}}}\)
\(\displaystyle{\sqrt[{{3}}]{{\lim_{{{x}\rightarrow\infty}}{\left({1}-\lim_{{{x}\rightarrow\infty}}{\left(\frac{{2}}{{x}}+\lim_{{{x}\rightarrow\infty}}{\left(\frac{{3}}{{x}^{{3}}}\right)}\right)}\right)}}}}\)
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({1}\right)}={1}\)
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left(\frac{{2}}{{x}}\right)}={0}\)
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left(\frac{{3}}{{x}^{{3}}}\right)}={0}\)
\(\displaystyle{\sqrt[{{3}}]{{{1}-{0}+{0}}}}={1}\)
The limit of the denominator is 2:
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({2}+\frac{{1}}{{x}}\right)}\)
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({2}\right)}+\lim_{{{x}\rightarrow\infty}}{\left(\frac{{1}}{{x}}\right)}\)
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({2}\right)}={2}\)
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left(\frac{{1}}{{x}}\right)}={0}\)
2+0=2
Simlifying all of this:
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({\left(\frac{{\sqrt[{{3}}]{{{x}^{{3}}-{2}{x}^{{2}}+{3}}}}}{{{2}{x}+{1}}}\right)}\right)}=\frac{{1}}{{2}}\)
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