Khaleesi Herbert

## Answered question

2021-02-25

Solve. $\underset{x\to \mathrm{\infty }}{lim}\left(\frac{\sqrt[3]{{x}^{3}-2{x}^{2}+3}}{2x+1}$

### Answer & Explanation

faldduE

Skilled2021-02-26Added 109 answers

$\underset{x\to \mathrm{\infty }}{lim}\left(\left(\frac{\sqrt[3]{{x}^{3}-2{x}^{2}+3}}{2x+1}\right)=\frac{1}{2}$
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:
$\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt[3]{1-\frac{2}{x}+\frac{3}{{x}^{3}}}\right)$
$\sqrt[3]{\underset{x\to \mathrm{\infty }}{lim}\left(1-\frac{2}{x}+\frac{3}{{x}^{3}}\right)}$
$\sqrt[3]{\underset{x\to \mathrm{\infty }}{lim}\left(1-\underset{x\to \mathrm{\infty }}{lim}\left(\frac{2}{x}+\underset{x\to \mathrm{\infty }}{lim}\left(\frac{3}{{x}^{3}}\right)\right)\right)}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?