29. $limlimits_{x \to infty} \frac{(2x - 5)^3(3x - 1)^2}{(2x - 1)^2(3x - 5)^3}$ is\n(a) $\frac{3}{5}$ (b)…

Explanation:

Step1: Analyze the limit of the numerator and denominator for problem 29

For (\lim_{x\rightarrow\infty}\frac{(2x - 5)^{3}(3x - 1)^{2}}{(2x - 1)^{2}(3x - 5)^{3}}), divide each term in the numerator and denominator by (x^{5}) (since the degree of the numerator ((3 + 2=5)) and the degree of the denominator ((2+3 = 5))). [ \begin{align*} \lim_{x\rightarrow\infty}\frac{(2x - 5)^{3}(3x - 1)^{2}}{(2x - 1)^{2}(3x - 5)^{3}}&=\lim_{x\rightarrow\infty}\frac{(2-\frac{5}{x})^{3}(3-\frac{1}{x})^{2}}{(2-\frac{1}{x})^{2}(3-\frac{5}{x})^{3}} \end{align*} ]

Step2: Evaluate the limit

As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{1}{x}=0). [ \begin{align*} \lim_{x\rightarrow\infty}\frac{(2-\frac{5}{x})^{3}(3-\frac{1}{x})^{2}}{(2-\frac{1}{x})^{2}(3-\frac{5}{x})^{3}}&=\frac{(2 - 0)^{3}(3-0)^{2}}{(2 - 0)^{2}(3-0)^{3}}\ &=\frac{2^{3}\times3^{2}}{2^{2}\times3^{3}}\ &=\frac{2}{3} \end{align*} ]

Step3: Analyze the limit for problem 30

For (\lim_{x\rightarrow-\infty}\frac{2^{x}+\cos x}{x^{2}+\sin x}), we know that (\lim_{x\rightarrow-\infty}2^{x}=0) (since (y = 2^{x}) is an exponential function with base (a = 2>1) and for (x\rightarrow-\infty), (2^{x}=\frac{1}{2^{-x}}\rightarrow0)), and (- 1\leqslant\cos x\leqslant1), (-1\leqslant\sin x\leqslant1). Also, (\lim_{x\rightarrow-\infty}x^{2}=\infty) [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{2^{x}+\cos x}{x^{2}+\sin x}&=\frac{\lim_{x\rightarrow-\infty}(2^{x}+\cos x)}{\lim_{x\rightarrow-\infty}(x^{2}+\sin x)}\ &=\frac{0 + c}{ \infty+ d}\quad(-1\leqslant c\leqslant1,- 1\leqslant d\leqslant1) \end{align*} ]

Answer:

For problem 29: B. (\frac{2}{3}) For problem 30: B. (0)