1.
Question 1. Using the graph of \(f(x)\), what is \(\lim_{x \to -2} f(x)\)?
From the graph, as \(x\) approaches \(-2\) from the left, \(f(x)\) decreases without bound. As \(x\) approaches \(-2\) from the right, \(f(x)\) also decreases without bound. Since both sides approach \(-\infty\), \(\lim_{x \to -2} f(x)=-\infty\). Answer
2.
Question 2. \(\lim_{x \to 4}\frac{4-x}{\sqrt{x}-2}=\)
Rationalize the denominator: \(\frac{4-x}{\sqrt{x}-2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}=\frac{(4-x)(\sqrt{x}+2)}{x-4}\). Since \(4-x=-(x-4)\), the expression becomes \(-\left(\sqrt{x}+2\right)\). Now substitute \(x=4\): \(-\left(\sqrt{4}+2\right)=-(2+2)=-4\). Answer
3.
Question 3. \(\lim_{x \to \infty}\frac{\sqrt[3]{27x^3}}{3x}=\)
Simplify the numerator: \(\sqrt[3]{27x^3}=\sqrt[3]{27}\sqrt[3]{x^3}=3x\) for \(x\to\infty\). Then \(\frac{\sqrt[3]{27x^3}}{3x}=\frac{3x}{3x}=1\). Therefore, the limit is \(1\). Answer
4.
Question 4. \(\lim_{x \to -3}\frac{8}{x+3}=\)
Check the one-sided limits. As \(x\to -3^-\), \(x+3\to 0^-\), so \(\frac{8}{x+3}\to -\infty\). As \(x\to -3^+\), \(x+3\to 0^+\), so \(\frac{8}{x+3}\to \infty\). Since the one-sided limits are not equal, the two-sided limit does not exist. Answer
5.
Question 5. True or False: \(\lim_{x \to \infty} e^{-\frac{2}{x}}=1\)
As \(x\to\infty\), \(-\frac{2}{x}\to 0\). Therefore, \(e^{-\frac{2}{x}}\to e^0=1\). The statement is true. Answer
6.
Question 6. If \(\lim_{x \to b} f(x)=2\) and \(\lim_{x \to b} g(x)=10\), then \(\lim_{x \to b}\left[3f(x)-4g(x)\right]=-34\).
Use the limit laws: \(\lim_{x \to b}\left[3f(x)-4g(x)\right]=3\lim_{x \to b}f(x)-4\lim_{x \to b}g(x)\). Substitute the given limits: \(3(2)-4(10)=6-40=-34\). The statement is true. Answer
7.
Question 7. \(\lim_{x \to \infty}\frac{3x-2}{x^3-1}=\)
Divide the numerator and denominator by the highest power in the denominator, \(x^3\): \(\frac{3x-2}{x^3-1}=\frac{\frac{3}{x^2}-\frac{2}{x^3}}{1-\frac{1}{x^3}}\). As \(x\to\infty\), \(\frac{3}{x^2}\to 0\), \(\frac{2}{x^3}\to 0\), and \(\frac{1}{x^3}\to 0\). Therefore, the limit is \(\frac{0-0}{1-0}=0\). Answer
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