<s>[INST] 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.
 [/INST] 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
$$\lim_{x\to0}\dfrac{\dfrac13\cdot\dfrac1{(1+x)^{\frac23}}+\dfrac13\cdot\dfrac1{(1-x)^{\frac23}}}{\sqrt[6]{1+x}+x\cdot\dfrac16\cdot\dfrac{1}{(1+x)^{\frac56}}+\sqrt[6]{1-x}-x\cdot\dfrac16\cdot\dfrac{1}{(1-x)^{\frac56}}}=\dfrac{\frac13+\frac13}{1+1}=\dfrac13$$
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