What is the answer for the limit with cube root?

Question

I have to calculate the limit $\lim_{x\to\infty}\dfrac{\sqrt[3]{x+3x^2}}{x+2}$

I have tried to solve it:

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

But I don't know what to do after. Thanks in advance

Answer 1

The highest power in both the numerator is $1$, i.e., $x^{1}$ in the denominator. We ignore the $x^2$ in the numerator since it is essentially $x^{2/3}$ because of the cube root.

Divide by $x^{1} = x$: for $x \neq 0$: $$\dfrac{\sqrt[3]{x+3x^2}}{x+2} = \dfrac{\sqrt[3]{x+3x^2}/x}{(x+2)/x} = \dfrac{\sqrt[3]{x+3x^2}/\sqrt[3]{x^3}}{1+2/x} = \dfrac{\sqrt[3]{(x+3x^2)/x^3}}{1+2/x} = \dfrac{\sqrt[3]{x^{-2}+3x^{-1}}}{1+2/x}\text{.}$$ Now $$\lim_{x \to \infty}\sqrt[3]{x^{-2}+3x^{-1}} = \sqrt[3]{\lim_{n \to \infty}\left(x^{-2}+3x^{-1}\right)} = \sqrt[3]{0+0} = 0 $$ and $$\lim_{x \to \infty}(1+2/x) = 1 + 0 = 1$$ hence $$\lim\limits_{x \to \infty}\dfrac{\sqrt[3]{x+3x^2}}{x+2} = \dfrac{0}{1} = 0\text{.}$$

Answer 2

We have $$\dfrac{\sqrt[3]{x+3x^2}}{x+2} =\sqrt[3]{\frac{x+3x^2}{(x+2)^3}} =\sqrt[3]{\frac{\frac1{x^2}+\frac3x}{(1+\frac2x)^3}}\ .$$ Can you now see what happens as $x\to\infty$?

Answer 3

Instead factor $x^{2/3}$ out of the numerator (and denominator):

$$ = \lim\limits_{x \to \infty} \dfrac{x^{2/3}}{x^{2/3}} \cdot \dfrac{\left(\frac{1}{x}+3\right)^{1/3}}{x^{1/3}+\frac{2}{x^{2/3}}} = \lim\limits_{x \to \infty} \dfrac{\left(\frac{1}{x}+3\right)^{1/3}}{x^{1/3}+\frac{2}{x^{2/3}}} $$

As $x\to\infty$, you'll get $\dfrac{(0+3)^{1/3}}{\infty+0} = 0$

Answer 4

$\lim_{n \rightarrow \infty} \dfrac{(x+3x^2)^{\frac{1}{3}}}{x+2} = \lim_{n \rightarrow \infty} \dfrac{(x^3(\dfrac{1}{x^2}+\dfrac{3}{x}))^{\frac{1}{3}}}{x(1+\dfrac{2}{x})} = \lim_{n \rightarrow \infty} \dfrac{(\dfrac{1}{x^2}+\dfrac{3}{x})^{\frac{1}{3}}}{(1+\dfrac{2}{x})} = 0$

Answer 5

Apply L'Hopital's rule, which states that the limit of a ratio (if both the numerator and denominator converge to 0 or $\infty$) is equal to the ratio of the derivatives. Then $\underset{x\to \infty}{\lim} \frac{(x+3x^2)^{1/3}}{x+2} =\underset{x\to \infty}{\lim} \frac{\frac{6 x+1}{3 \left(3 x^2+x\right)^{2/3}}}{1} =0. $