Question

determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
lim_{x
ightarrowinfty}\frac{(3x - 2)(6 + 5x^{3})}{(1 - 7x^{2})(5x + 6)}

Explanation:

Step1: Expand the numerator and denominator

Expand \((3x - 2)(6 + 5x^{3})\) and \((1-7x^{2})(5x + 6)\).
\((3x - 2)(6 + 5x^{3})=3x\times6+3x\times5x^{3}-2\times6 - 2\times5x^{3}=18x + 15x^{4}-12-10x^{3}=15x^{4}-10x^{3}+18x - 12\)
\((1 - 7x^{2})(5x + 6)=1\times5x+1\times6-7x^{2}\times5x-7x^{2}\times6=5x + 6-35x^{3}-42x^{2}=-35x^{3}-42x^{2}+5x + 6\)

Step2: Divide each term by the highest - power of \(x\) in the denominator (\(x^{3}\))

\(\lim_{x
ightarrow\infty}\frac{15x^{4}-10x^{3}+18x - 12}{-35x^{3}-42x^{2}+5x + 6}=\lim_{x
ightarrow\infty}\frac{\frac{15x^{4}}{x^{3}}-\frac{10x^{3}}{x^{3}}+\frac{18x}{x^{3}}-\frac{12}{x^{3}}}{\frac{-35x^{3}}{x^{3}}-\frac{42x^{2}}{x^{3}}+\frac{5x}{x^{3}}+\frac{6}{x^{3}}}\)
\(=\lim_{x
ightarrow\infty}\frac{15x - 10+\frac{18}{x^{2}}-\frac{12}{x^{3}}}{-35-\frac{42}{x}+\frac{5}{x^{2}}+\frac{6}{x^{3}}}\)

Step3: Evaluate the limit

As \(x
ightarrow\infty\), \(\frac{18}{x^{2}}
ightarrow0\), \(\frac{12}{x^{3}}
ightarrow0\), \(\frac{42}{x}
ightarrow0\), \(\frac{5}{x^{2}}
ightarrow0\) and \(\frac{6}{x^{3}}
ightarrow0\)
\(\lim_{x
ightarrow\infty}\frac{15x - 10+\frac{18}{x^{2}}-\frac{12}{x^{3}}}{-35-\frac{42}{x}+\frac{5}{x^{2}}+\frac{6}{x^{3}}}=\infty\) since the numerator goes to \(\infty\) and the denominator is a non - zero constant (\(- 35\)) as \(x
ightarrow\infty\)

Answer:

The limit does not exist (DNE)