Evaluate $\lim_{x\to\infty} ((x^5+x^4)^{1/6}-(x^5-x^4)^{1/6})$ I've been struggling with the following:
$$\lim_{x\to\infty} ((x^6+x^5)^{1/6}-(x^6-x^5)^{1/6})$$
Tried factoring out $x^{5/6}$ and then using L'hopital- which got me nowhere, tried multiplying by the conjugate, but it got messy- so either I'm scared of the algebra or there's a better way.
 A: When you have indeterminate type of form $\infty-\infty$ you need to rationalize your expression to get fraction:
$$\lim_{x\to\infty}\left(\sqrt[6]{x^5+x^4}-\sqrt[6]{x^5-x^4}\right)=\lim_{x\to\infty}\dfrac{\sqrt[3]{x^5+x^4}-\sqrt[3]{x^5-x^4}}{\sqrt[6]{x^5+x^4}+\sqrt[6]{x^5-x^4}}=\lim_{x\to0}\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x\sqrt[6]{1+x}+x\sqrt[6]{1-x}}$$
Now apply L'Hopital's rule
$$\lim_{x\to0}\dfrac{\dfrac13\cdot\dfrac1{(1+x)^{\frac23}}+\dfrac13\cdot\dfrac1{(1-x)^{\frac23}}}{\sqrt[6]{1+x}+x\cdot\dfrac16\cdot\dfrac{1}{(1+x)^{\frac56}}+\sqrt[6]{1-x}-x\cdot\dfrac16\cdot\dfrac{1}{(1-x)^{\frac56}}}=\dfrac{\frac13+\frac13}{1+1}=\dfrac13$$
A: One more way (without L'Hospital):
After getting 
$$
\lim_{x \to 0} \frac{(1+x)^{\frac{1}{3}} - (1-x)^{\frac{1}{3}}}{x((1+x)^{\frac{1}{6}} + (1-x)^{\frac{1}{6}})}
$$
multiply both numerator and denominator by $(1+x)^{\frac{2}{3}}  +(1-x)^{\frac{2}{3}} + ((1+x)(1-x))^{\frac{1}{3}}$ to get $a^3 -b^3$ in the numerator:
$$
\lim_{x \to 0} \frac{2}{((1+x)^{\frac{1}{6}} + (1-x)^{\frac{1}{6}})((1+x)^{\frac{2}{3}}  +(1-x)^{\frac{2}{3}} + ((1+x)(1-x))^{\frac{1}{3}})}
$$
Now take the limit to get the result. 
