1.
Question 1. What is the \(\lim_{x\to 1^+} f(x)\) for the function \(f(x)\) graphed below?
We only care about the values that are greater than 1. Looking at the right side of the graph, the function approaches \(2\). Therefore, \(\lim_{x\to 1^+} f(x)=2\). Answer: D
2.
Question 2. \(\lim_{x\to -2}\frac{x^2-4}{x+2}=\)
Factor the numerator: \(x^2-4=(x-2)(x+2)\). Then \(\frac{x^2-4}{x+2}=\frac{(x-2)(x+2)}{x+2}=x-2\), for \(x\ne -2\). Now substitute \(x=-2\): \(-2-2=-4\). Answer: A
3.
Question 3. \(\lim_{x\to \infty}\frac{\sqrt[3]{8x^3}}{2x}=\)
Simplify the cube root: \(\sqrt[3]{8x^3}=2x\). Then \(\frac{\sqrt[3]{8x^3}}{2x}=\frac{2x}{2x}=1\). Therefore, the limit is \(1\). Answer: E
4.
Question 4. What is \(\lim_{x\to -5^-}\frac{x}{x+5}\)?
As \(x\to -5^-\), the numerator approaches \(-5\), and the denominator \(x+5\) approaches \(0^-\). A negative number divided by a very small negative number approaches \(\infty\). Therefore, the limit is \(\infty\). Answer: C
5.
Question 5. True or False: \(\lim_{x\to 0^-} e^{3/x}=0\).
As \(x\to 0^-\), \(\frac{3}{x}\to -\infty\). Therefore, \(e^{3/x}\to e^{-\infty}=0\). The statement is true. Answer: A
6.
Question 6. If \(\lim_{x\to b} f(x)=5\) and \(\lim_{x\to b} g(x)=10\), then \(\lim_{x\to b}\left[3f(x)-4g(x)\right]=-25\).
Use 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: \(3(5)-4(10)=15-40=-25\). The statement is true. Answer: B
7.
Question 7. \(\lim_{x\to -\infty}\frac{14x^4-4x}{17x^3-7x^4}=\)
Divide numerator and denominator by \(x^4\): \(\frac{14-\frac{4}{x^3}}{\frac{17}{x}-7}\). As \(x\to -\infty\), \(\frac{4}{x^3}\to 0\) and \(\frac{17}{x}\to 0\). So the limit is \(\frac{14}{-7}=-2\). Answer: E
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