Question

# The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: Divide nu

Rational functions
The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits.
$$\displaystyle\lim_{{{x}\rightarrow-\infty}}{\frac{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}-{5}{x}+{3}}}{{{2}{x}+{x}^{{{\frac{{{2}}}{{{3}}}}}}-{4}}}}$$

$$\displaystyle\lim_{{{x}\rightarrow-\infty}}{\frac{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}-{5}{x}+{3}}}{{{2}{x}+{x}^{{{\frac{{{2}}}{{{3}}}}-{4}}}}}}:{\frac{{{x}}}{{{x}}}}=\lim_{{{x}\rightarrow-\infty}}{\frac{{{x}^{{-{\frac{{{2}}}{{{3}}}}-{5}+{\frac{{{3}}}{{{x}}}}}}}}{{{2}+{x}^{{-{\frac{{{1}}}{{{3}}}}-{\frac{{{4}}}{{{x}}}}}}}}}$$=
Use that $$\displaystyle\lim_{{{x}\rightarrow-\infty}}{x}^{{-{n}}}={0}$$.
$$\displaystyle={\frac{{{0}-{5}+{0}}}{{{2}+{0}-{0}}}}=-{\frac{{{5}}}{{{2}}}}$$