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{-4(9x - 1)}{(4 - 7x^{2})(8 + 3x^{2})}

Explanation:

Step1: Expand the denominator

First, expand \((4 - 7x^{2})(8 + 3x^{2})\) using the FOIL - method. \((4 - 7x^{2})(8 + 3x^{2})=4\times8+4\times3x^{2}-7x^{2}\times8 - 7x^{2}\times3x^{2}=32 + 12x^{2}-56x^{2}-21x^{4}=32 - 44x^{2}-21x^{4}\). So the function becomes \(\frac{-4(9x - 1)}{32 - 44x^{2}-21x^{4}}=\frac{-36x + 4}{32 - 44x^{2}-21x^{4}}\).

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

The highest - power of \(x\) in the denominator is \(x^{4}\). Divide each term in the numerator and denominator by \(x^{4}\): \(\lim_{x
ightarrow\infty}\frac{\frac{-36x}{x^{4}}+\frac{4}{x^{4}}}{\frac{32}{x^{4}}-\frac{44x^{2}}{x^{4}}-\frac{21x^{4}}{x^{4}}}=\lim_{x
ightarrow\infty}\frac{-\frac{36}{x^{3}}+\frac{4}{x^{4}}}{\frac{32}{x^{4}}-\frac{44}{x^{2}}-21}\).

Step3: Evaluate the limit of each term

We know that \(\lim_{x
ightarrow\infty}\frac{c}{x^{n}} = 0\) for any non - zero constant \(c\) and positive integer \(n\). So, \(\lim_{x
ightarrow\infty}-\frac{36}{x^{3}} = 0\), \(\lim_{x
ightarrow\infty}\frac{4}{x^{4}} = 0\), \(\lim_{x
ightarrow\infty}\frac{32}{x^{4}} = 0\), and \(\lim_{x
ightarrow\infty}-\frac{44}{x^{2}} = 0\).
Then \(\lim_{x
ightarrow\infty}\frac{-\frac{36}{x^{3}}+\frac{4}{x^{4}}}{\frac{32}{x^{4}}-\frac{44}{x^{2}}-21}=\frac{0 + 0}{0 - 0-21}\).

Answer:

\(0\)