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
ANSWERED
asked 2021-06-13
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.
\(\displaystyle\lim_{{{x}\rightarrow-\infty}}{\frac{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}-{5}{x}+{3}}}{{{2}{x}+{x}^{{{\frac{{{2}}}{{{3}}}}}}-{4}}}}\)

Answers (1)

2021-06-14
Highest power of x in denominator is x. Divide both numerator and denominator with x.
\(\displaystyle\lim_{{{x}\rightarrow-\infty}}{\frac{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}-{5}{x}+{3}}}{{{2}{x}+{x}^{{{\frac{{{2}}}{{{3}}}}-{4}}}}}}:{\frac{{{x}}}{{{x}}}}=\lim_{{{x}\rightarrow-\infty}}{\frac{{{x}^{{-{\frac{{{2}}}{{{3}}}}-{5}+{\frac{{{3}}}{{{x}}}}}}}}{{{2}+{x}^{{-{\frac{{{1}}}{{{3}}}}-{\frac{{{4}}}{{{x}}}}}}}}}\)=
Use that \(\displaystyle\lim_{{{x}\rightarrow-\infty}}{x}^{{-{n}}}={0}\).
\(\displaystyle={\frac{{{0}-{5}+{0}}}{{{2}+{0}-{0}}}}=-{\frac{{{5}}}{{{2}}}}\)
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