Solved: "Solve. limx→∞(x3−2x2+332x+1"

Khaleesi Herbert

Answered question

2021-02-25

Solve.

lim_{x \to \infty} (\frac{\sqrt[3]{x^{3} - 2 x^{2} + 3}}{2 x + 1}

faldduE

Skilled2021-02-26Added 109 answers

lim_{x \to \infty} ((\frac{\sqrt[3]{x^{3} - 2 x^{2} + 3}}{2 x + 1}) = \frac{1}{2}

To start, divide by the highest denominator power by using the following algebraic property:

a + b = a (1 + \frac{b}{a})

(\sqrt[3]{x^{3} - 2 x^{2} + 3} = (\sqrt[3]{x^{3} (1 - \frac{2}{x} + \frac{3}{x^{3}})}

x \to \infty \Rightarrow \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}}}}{2 x + 1}

Then, divide by x and refine:

\frac{\sqrt[3]{1 - \frac{2}{x} + \frac{3}{x^{3}}}}{x} / (2 \frac{x}{2} + \frac{1}{x})

= \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:

lim_{x \to \infty} [(\frac{f (x)}{g (x)})] = \frac{lim_{x \to \infty} f (x)}{lim_{x \to \infty} g (x)}, lim_{x \to \infty} g (x)

\frac{lim (x \to \infty) (\sqrt[3]{1 - \frac{2}{x} + \frac{3}{x^{3}}})}{lim_{x \to \infty} (2 + \frac{1}{x})}

The limit of the numerator is 1:

lim_{x \to \infty} (\sqrt[3]{1 - \frac{2}{x} + \frac{3}{x^{3}}})

\sqrt[3]{lim_{x \to \infty} (1 - \frac{2}{x} + \frac{3}{x^{3}})}

\sqrt[3]{lim_{x \to \infty} (1 - lim_{x \to \infty} (\frac{2}{x} + lim_{x \to \infty} (\frac{3}{x^{3}})))}

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