evaluate the following limits. if needed, enter inf for ∞ and -in (a) lim x→∞ (2 - x)(8 + 5x) / (3 - 3x)(11…

Explanation:

Step1: Expand the numerator and denominator

First, expand ((2 - x)(8 + 5x)=16 + 10x-8x - 5x^{2}=16 + 2x-5x^{2}) and ((3 - 3x)(11 + 10x)=33+30x - 33x-30x^{2}=33 - 3x-30x^{2}). So the function becomes (\lim_{x\rightarrow\pm\infty}\frac{16 + 2x-5x^{2}}{33 - 3x-30x^{2}}).

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

For (\lim_{x\rightarrow\pm\infty}\frac{16 + 2x-5x^{2}}{33 - 3x-30x^{2}}), divide each term by (x^{2}): (\lim_{x\rightarrow\pm\infty}\frac{\frac{16}{x^{2}}+\frac{2}{x}-5}{\frac{33}{x^{2}}-\frac{3}{x}-30}).

Step3: Evaluate the limit as (x\rightarrow\pm\infty)

As (x\rightarrow\pm\infty), (\lim_{x\rightarrow\pm\infty}\frac{16}{x^{2}} = 0), (\lim_{x\rightarrow\pm\infty}\frac{2}{x}=0), (\lim_{x\rightarrow\pm\infty}\frac{33}{x^{2}} = 0) and (\lim_{x\rightarrow\pm\infty}\frac{3}{x}=0). Then (\lim_{x\rightarrow\pm\infty}\frac{\frac{16}{x^{2}}+\frac{2}{x}-5}{\frac{33}{x^{2}}-\frac{3}{x}-30}=\frac{0 + 0-5}{0 - 0-30}).

Answer:

(\frac{1}{6})