<s>[INST] verify this limit of radical function $$\lim_{x \to -\infty} \frac{\sqrt[3]{x}- \sqrt[5]{x}}{4\sqrt[3]{x}+\sqrt[5]{x}} $$  
$$= \lim_{x \to -\infty} \frac{x^{1/3}-x^{1/5}}{4x^{1/3}+x^{1/5}} $$ 
divide numerator and denominator by highest power of $x$ in the denominator, $x^{1/3},$ to get:
$$= \lim_{x \to -\infty} \frac{1 - x^{-2/15}}{4+x^{-2/15}}$$ 
Which gives an answer of $\frac{1}{4}.$ Is this correct? Did I show work correctly?
 [/INST] Your work is correct
$$\lim_{x\to-\infty}\frac{\sqrt[3]{x}-\sqrt[5]{x}}{4\sqrt[3]{x}+\sqrt[5]{x}}=$$ 
$$\lim_{x\to-\infty}\frac{\sqrt[15]{x^5}-\sqrt[15]{x^3}}{4\sqrt[15]{x^5}+\sqrt[15]{x^3}}=$$
$$\lim_{x\to-\infty}\frac{\frac{\sqrt[15]{x^5}-\sqrt[15]{x^3}}{\sqrt[15]{x^5}}}{4\frac{\sqrt[15]{x^5}+\sqrt[15]{x^3}}{\sqrt[15]{x^5}}}=$$ 
$$\lim_{x\to-\infty}\frac{1-\sqrt[15]{1/x^2}}{4(1+\sqrt[15]{1/x^2})}=1/4$$   
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