Answer
$$\lim_{x\to-\infty}\frac{\sqrt[3]x-\sqrt[5]x}{\sqrt[3]x+\sqrt[5]x}=1$$
Work Step by Step
We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator.
Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$
$$A=\lim_{x\to-\infty}\frac{\sqrt[3]x-\sqrt[5]x}{\sqrt[3]x+\sqrt[5]x}=\lim_{x\to-\infty}\frac{x^{1/3}-x^{1/5}}{x^{1/3}+x^{1/5}}$$
The highest power of $x$ in the denominator here is $x^{1/3}$, so we divide both numerator and denominator by $x^{1/3}$: $$A=\lim_{x\to-\infty}\frac{1-x^{1/5-1/3}}{1+x^{1/5-1/3}}=\lim_{x\to-\infty}\frac{1-x^{-2/15}}{1+x^{-2/15}}=\lim_{x\to-\infty}\frac{1-\frac{1}{x^{2/15}}}{1+\frac{1}{x^{2/15}}}$$ $$A=\frac{1-0}{1+0}=\frac{1}{1}=1$$