26. $limlimits_{x \to 2}\frac{(x - 2)(x + 1)}{x^{2}-x - 2}$ is\n(a) $-1$\n(b) $1$\n(c) $2$\n(d)…

26. $limlimits_{x \to 2}\frac{(x - 2)(x + 1)}{x^{2}-x - 2}$ is\n(a) $-1$\n(b) $1$\n(c) $2$\n(d) nonexistent\n27. $limlimits_{x \to -infty}\frac{2 - x - x^{2}}{sqrt3{8x^{6}+2x^{4}+1}}$ is\n(a) $-infty$\n(b) $-\frac{1}{2}$\n(c) $0$\n(d) $\frac{1}{2}$\n28. $limlimits_{x \to 0}\frac{sqrt{x^{2}+pi^{2}}+pi}{x^{2}}$ is\n(a) $-infty$\n(b) $0$\n(c) $2pi$\n(d) $infty$

26. $limlimits_{x \to 2}\frac{(x - 2)(x + 1)}{x^{2}-x - 2}$ is\n(a) $-1$\n(b) $1$\n(c) $2$\n(d) nonexistent\n27. $limlimits_{x \to -infty}\frac{2 - x - x^{2}}{sqrt3{8x^{6}+2x^{4}+1}}$ is\n(a) $-infty$\n(b) $-\frac{1}{2}$\n(c) $0$\n(d) $\frac{1}{2}$\n28. $limlimits_{x \to 0}\frac{sqrt{x^{2}+pi^{2}}+pi}{x^{2}}$ is\n(a) $-infty$\n(b) $0$\n(c) $2pi$\n(d) $infty$

Answer

Problem 26

Explanation:

Step1: Factor the denominator

Factor (x^{2}-x - 2=(x - 2)(x+1))

Step2: Simplify the function

(\lim_{x\rightarrow2}\frac{(x - 2)(x + 1)}{x^{2}-x - 2}=\lim_{x\rightarrow2}\frac{(x - 2)(x + 1)}{(x - 2)(x + 1)}) Since (x\rightarrow2) (not equal to (2) when taking the limit), we can cancel ((x - 2)(x + 1)) (for (x\neq2) and (x\neq - 1)), and the function simplifies to (1)

Answer:

B. (1)

Problem 27

Explanation:

Step1: Divide numerator and denominator by (|x|^{2})

For (x\rightarrow-\infty), (|x|=-x). The numerator (2 - x - x^{2}=-x^{2}(1+\frac{1}{x}-\frac{2}{x^{2}})) and the denominator (\sqrt[3]{8x^{6}+2x^{4}+1}=\sqrt[3]{x^{6}(8 + \frac{2}{x^{2}}+\frac{1}{x^{6}})}=-x^{2}\sqrt[3]{8+\frac{2}{x^{2}}+\frac{1}{x^{6}}}) (because (x<0) as (x\rightarrow-\infty))

Step2: Calculate the limit

(\lim_{x\rightarrow-\infty}\frac{2 - x - x^{2}}{\sqrt[3]{8x^{6}+2x^{4}+1}}=\lim_{x\rightarrow-\infty}\frac{-x^{2}(1+\frac{1}{x}-\frac{2}{x^{2}})}{-x^{2}\sqrt[3]{8+\frac{2}{x^{2}}+\frac{1}{x^{6}}}}) Cancel out (-x^{2}) (since (x\neq0) as (x\rightarrow-\infty)), and use (\lim_{x\rightarrow-\infty}\frac{1}{x}=0,\lim_{x\rightarrow-\infty}\frac{1}{x^{2}} = 0,\lim_{x\rightarrow-\infty}\frac{1}{x^{6}}=0) (\lim_{x\rightarrow-\infty}\frac{1+\frac{1}{x}-\frac{2}{x^{2}}}{\sqrt[3]{8+\frac{2}{x^{2}}+\frac{1}{x^{6}}}}=\frac{1 + 0-0}{\sqrt[3]{8+0 + 0}}=\frac{1}{2})

Answer:

D. (\frac{1}{2})

Problem 28

Explanation:

Step1: Analyze the behavior of the function near (x = 0)

As (x\rightarrow0), (\sqrt{x^{2}+\pi^{2}}+\pi=\pi\sqrt{1+\frac{x^{2}}{\pi^{2}}}+\pi\approx\pi(1+\frac{x^{2}}{2\pi^{2}})+\pi) (using the binomial approximation (\sqrt{1 + t}\approx1+\frac{t}{2}) for (|t|\ll1), here (t=\frac{x^{2}}{\pi^{2}})) (\sqrt{x^{2}+\pi^{2}}+\pi\approx2\pi+\frac{x^{2}}{2\pi})

Step2: Calculate the limit

(\lim_{x\rightarrow0}\frac{\sqrt{x^{2}+\pi^{2}}+\pi}{x^{2}}=\lim_{x\rightarrow0}\frac{2\pi+\frac{x^{2}}{2\pi}}{x^{2}}) (\lim_{x\rightarrow0}(\frac{2\pi}{x^{2}}+\frac{1}{2\pi})=\infty)

Answer:

D. (\infty)