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

Question

I really don't know what to try to solve this limit.

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

If I multiply by x^2, it becomes an indeterminate form: + ∞ - ∞

If instead I simplify $1-\sqrt[3]{\frac{\left(x^2-1\right)}{\left(x^2+1\right)}}$ I get to ∞ * 0.

So what could I try? I don't see a way to rewrite $x^2+1$

Answer 1

Let $$t=\sqrt[3]{\frac{\left(x^2-1\right)}{\left(x^2+1\right)}}\implies x^2=-{t^3+1\over t^3-1}$$ then

$$\lim _{t\to 1}-{t^3+1\over t^3-1}(1-t )=\lim _{t\to1 }{t^3+1\over t^2+t+1}={2\over 3}$$

Answer 2

Write $$\frac{\sqrt[3]{x^2+1}-\sqrt[3]{x^2-1}}{\sqrt[3]{x^2+1}}$$ and multiply numerator and denominator by $$\sqrt[3]{x^2+1}^2+\sqrt{x^2+1}\sqrt[3]{x^2-1}+\sqrt[3]{x^2-1}^2$$

Answer 3

As usual,

$$1-\sqrt[3]z=\frac{1-z}{1+\sqrt[3]z+\sqrt[3]z^2}$$ and with

$$z:=\frac{x^2-1}{x^2+1}$$

the denominator tends to $3$ so that the limit reduces to that of

$$\frac13x^2\left(1-\frac{x^2-1}{x^2+1}\right).$$

$$\frac23.$$

Answer 4

series method...
As $x \to \infty$: \begin{align} \frac{x^2 - 1}{x^2+1} &= 1 - \frac{2}{x^2} + O(x^{-4})\qquad\text{(long division)} \\ \left(\frac{x^2 - 1}{x^2+1}\right)^{1/3} &= 1 - \frac{2}{3x^2} + O(x^{-4}) \\ 1 - \left(\frac{x^2 - 1}{x^2+1}\right)^{1/3} &= \frac{2}{3x^2} + O(x^{-4}) \\ x^2\left(1 - \left(\frac{x^2 - 1}{x^2+1}\right)^{1/3}\right) &= \frac{2}{3} + O(x^{-2}) \end{align} and thus $$ \lim _{x\to \infty }x^2\left(1-\sqrt[3]{\frac{\left(x^2-1\right)}{\left(x^2+1\right)}}\;\right) = \frac{2}{3} $$

Answer 5

You may set $x^2=\frac 1t$ and consider $t\to 0^+$.

Then, you only need the derivatives

• $(\star)$: $(\sqrt[3]{1-t})' =-\frac{1}{3\sqrt[3]{(1-t)^2}}$ and $(\sqrt[3]{1+t})'=\frac{1}{3\sqrt[3]{(1+t)^2}}$:

\begin{eqnarray*}x^2\left(1-\sqrt[3]{\frac{\left(x^2-1\right)}{\left(x^2+1\right)}}\right) & \stackrel{x^2=\frac 1t}{=} & \frac{1-\sqrt[3]{\frac{1-t}{1+t}}}{t}\\ & = & \frac{1}{\sqrt[3]{1+t}}\cdot \frac{\sqrt[3]{1+t} - \sqrt[3]{1-t}}{t} \\ & = & \frac{1}{\sqrt[3]{1+t}}\cdot \left(\frac{\sqrt[3]{1+t} -1 }{t} - \frac{\sqrt[3]{1-t} -1}{t}\right) \\ & \stackrel{t \to 0^+}{\longrightarrow} & \left. (\sqrt[3]{1+t})'\right|_{t=0} - \left. (\sqrt[3]{1-t})'\right|_{t=0}\\ & \stackrel{(\star)}{=} & \frac 13 + \frac 13 = \frac{2}{3} \end{eqnarray*}