hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)\nscore: 3/6 answered: 3/6\nquestion 4\nevaluate…

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

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

(\lim_{x\rightarrow\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\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{12}{x^{2}} = 0), (\lim_{x\rightarrow\infty}\frac{2}{x}=0), (\lim_{x\rightarrow\infty}\frac{3}{x^{2}} = 0), (\lim_{x\rightarrow\infty}\frac{30}{x}=0) So (\lim_{x\rightarrow\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})