Question

hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)
score: 3/6 answered: 3/6
question 4
evaluate the limit
lim_{x
ightarrowinfty}\frac{(2 - x)(6 + 4x)}{(3 - 3x)(1+ 11x)}

Explanation:

Step1: Expand the numerator and denominator

First, expand \((2 - x)(6 + 4x)\) and \((3-3x)(1 + 11x)\).
\((2 - x)(6 + 4x)=12+8x-6x - 4x^{2}=12 + 2x-4x^{2}\)
\((3-3x)(1 + 11x)=3+33x-3x-33x^{2}=3 + 30x-33x^{2}\)
So the limit becomes \(\lim_{x
ightarrow\infty}\frac{12 + 2x-4x^{2}}{3 + 30x-33x^{2}}\)

Step2: Divide numerator and denominator by \(x^{2}\)

\(\lim_{x
ightarrow\infty}\frac{\frac{12}{x^{2}}+\frac{2}{x}-4}{\frac{3}{x^{2}}+\frac{30}{x}-33}\)

Step3: Evaluate the limit of each term

As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{12}{x^{2}} = 0\), \(\lim_{x
ightarrow\infty}\frac{2}{x}=0\), \(\lim_{x
ightarrow\infty}\frac{3}{x^{2}} = 0\), \(\lim_{x
ightarrow\infty}\frac{30}{x}=0\)
So \(\lim_{x
ightarrow\infty}\frac{\frac{12}{x^{2}}+\frac{2}{x}-4}{\frac{3}{x^{2}}+\frac{30}{x}-33}=\frac{0 + 0-4}{0+0 - 33}=\frac{4}{33}\)

Answer:

\(\frac{4}{33}\)