1.
Question 1 Using the graph of \(f(x)\), what is the \(\lim_{x\to -2} f(x)\)?
Solution: From the graph, as \(x\) approaches \(-2\) from both the left and the right, \(f(x)\) decreases without bound. Therefore, \(\lim_{x\to -2} f(x)=-\infty\). Answer: \(-\infty\)
2.
Question 2 Evaluate \(\lim_{x\to -2}\frac{x^2-4}{x+2}\).
Solution: Factor the numerator: \(x^2-4=(x-2)(x+2)\). Then \(\lim_{x\to -2}\frac{x^2-4}{x+2}=\lim_{x\to -2}\frac{(x-2)(x+2)}{x+2}=\lim_{x\to -2}(x-2)=-2-2=-4\). Answer: -4
3.
Question 3 Evaluate \(\lim_{x\to \infty}\frac{\sqrt[3]{27x^3}}{3x}\).
Solution: Since \(\sqrt[3]{27x^3}=\sqrt[3]{27}\sqrt[3]{x^3}=3x\), we have \(\lim_{x\to \infty}\frac{\sqrt[3]{27x^3}}{3x}=\lim_{x\to \infty}\frac{3x}{3x}=1\). Answer: 1
4.
Question 4 Evaluate \(\lim_{x\to -3}\frac{8}{x+3}\).
Solution: 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 the same, the two-sided limit does not exist. Answer: Does not exist
5.
Question 5 Ture or False: \(\lim_{x\to \infty}e^{-\frac{2}{x}}=1\).
Solution: As \(x\to \infty\), \(-\frac{2}{x}\to 0\). Therefore, \(e^{-\frac{2}{x}}\to e^0=1\). The statement is true. Answer: Ture
6.
Question 6 If \(\lim_{x\to b} f(x)=5\) and \(\lim_{x\to b} g(x)=10\), then \(\lim_{x\to b}[3f(x)-4g(x)]=-25\).
Solution: Use the limit laws: \(\lim_{x\to b}[3f(x)-4g(x)]=3\lim_{x\to b}f(x)-4\lim_{x\to b}g(x)=3(5)-4(10)=15-40=-25\). The statement is true. Answer: Ture
7.
Question 7 Evaluate \(\lim_{x\to \infty}\frac{14x^4-4x}{17x^3-7x^4}\).
Solution: Divide the numerator and denominator by \(x^4\): \(\lim_{x\to \infty}\frac{14-\frac{4}{x^3}}{\frac{17}{x}-7}=\frac{14-0}{0-7}=-2\). Answer: -2
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