# Evaluate $\lim\limits_{x\to+\infty} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right )$?

How do you evaluate $$\lim\limits_{x\to+\infty} \sqrt{x}\left (\sqrt[3]{x+1}-\sqrt[3]{x-1}\right ) ?$$

Thanks in advance for your support.

• Multiply and divide by $(x+1)^{2/3} + (x+1)^{1/3}(x-1)^{1/3} + (x-1)^{2/3}$. – TenaliRaman Feb 23 at 13:30
• Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn Feb 23 at 13:47

## 4 Answers

Hint: multiply the numerator and the denominator for $$(x+1)^{2/3} + (x-1)^{2/3} + (x^2 -1)^{1/3}$$. What happens?

$$...=\lim\limits_{x\to\infty} {2\sqrt{x}\over \sqrt[3]{(x+1)^2}+\sqrt[3]{x^2-1}+\sqrt[3]{(x-1)^2}}$$ $$=\lim\limits_{x\to\infty} {2\sqrt{x}\over 3\sqrt[3]{x^2}}$$ $$={2\over 3}\lim\limits_{x\to\infty} {1\over \sqrt[6]{x}}$$ $$=0$$

Let $$1/x=h^6$$ as $$[2,3]=6$$

$$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h^6}-\sqrt[3]{1-h^6}}{h^{3+2}}$$

$$=\lim_{h\to0^+}\dfrac{1+h^6-(1-h^6)}{h^5}\cdot\lim_{h\to0^+}\dfrac1{(1+h^6)^{2/3}+(1+h^6)^{1/3}(1-h^6)^{1/3}+(1-h^6)^{2/3}}$$

$$=0\cdot\dfrac1{(1+1\cdot1+1)}$$

If it is ok for you to use only the fact that $$\color{blue}{f(t)} =\sqrt[3]{1+t}$$ and $$\color{blue}{g(t)}=\sqrt[3]{1-t}$$ are differentiable at $$t= 0$$, then without calculating any derivative you may proceed as follows:

• Set $$x = \frac{1}{t}$$ and consider $$t \rightarrow 0^+$$

$$\begin{eqnarray*} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right ) & \stackrel{x=\frac{1}{t}}{=} & \frac{\sqrt[3]{1+t} - \sqrt[3]{1-t}}{\underbrace{\sqrt[3]{t}\cdot \sqrt{t}}_{=t^{\frac{5}{6}}}}\\ & = & \frac{\sqrt[3]{1+t} -1 + 1 - \sqrt[3]{1-t}}{t}\cdot t^{\frac{1}{6}}\\ & = & \left(\underbrace{\frac{\sqrt[3]{1+t} -1}{t}}_{\stackrel{t\to 0^+}{\longrightarrow}\color{blue}{f'(0)}} - \underbrace{\frac{\sqrt[3]{1-t} - 1}{t}}_{\stackrel{t\to 0^+}{\longrightarrow}\color{blue}{g'(0)}}\right) \cdot t^{\frac{1}{6}}\\ & \stackrel{t\to 0^+}{\longrightarrow} & (f'(0) - g'(0))\cdot 0 = 0 \end{eqnarray*}$$