1.
Question 1 The limit \(\lim_{x\to 2} f(x)\) does not exist for the following function, \(f(x)\).
Solution: From the graph, as \(x\) approaches \(2\) from the left, \(f(x)\) approaches \(2\). As \(x\) approaches \(2\) from the right, \(f(x)\) also approaches \(2\). The point at \(x=2\) is open, so the function is not continuous there, but the left-hand and right-hand limits are the same. Therefore, \(\lim_{x\to 2} f(x)\) exists. The statement that the limit does not exist is false. Answer: False
2.
Question 2 \(\lim_{x\to 4} \frac{4-x}{\sqrt{x}-2}=\)
Solution: Direct substitution gives \(\frac{4-4}{\sqrt{4}-2}=\frac{0}{0}\), so simplify first. Multiply by the conjugate: \(\frac{4-x}{\sqrt{x}-2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}=\frac{(4-x)(\sqrt{x}+2)}{x-4}\). Since \(4-x=-(x-4)\), this becomes \(-\left(\sqrt{x}+2\right)\). Now substitute \(x=4\): \(-\left(\sqrt{4}+2\right)=-(2+2)=-4\). Answer: \(-4\)
3.
Question 3 \(\lim_{x\to \infty} \frac{\sqrt[3]{8x^3}}{2x}=\)
Solution: Simplify the numerator: \(\sqrt[3]{8x^3}=\sqrt[3]{8}\sqrt[3]{x^3}=2x\). Therefore, \(\frac{\sqrt[3]{8x^3}}{2x}=\frac{2x}{2x}=1\) for large positive \(x\). Thus \(\lim_{x\to \infty} \frac{\sqrt[3]{8x^3}}{2x}=1\). Answer: \(1\)
4.
Question 4 \(\lim_{x\to 3^+} \frac{x}{x-3}=\)
Solution: As \(x\to 3^+\), the numerator \(x\) approaches \(3\), which is positive. The denominator \(x-3\) approaches \(0\) from the positive side, written \(0^+\). A positive number divided by a very small positive number increases without bound: \(\frac{3}{0^+}=\infty\). Answer: \(\infty\)
5.
Question 5 Evaluate \(\lim_{x\to 0^-} e^{\frac{3}{x}}=\)
Solution: As \(x\to 0^-\), \(x\) is negative and very close to \(0\), so \(\frac{3}{x}\to -\infty\). Therefore, \(e^{\frac{3}{x}}\to e^{-\infty}=0\). Answer: \(0\)
6.
Question 6 If \(\lim_{x\to b} f(x)=2\) and \(\lim_{x\to b} g(x)=10\), then \(\lim_{x\to b} [3f(x)-4g(x)]=?\)
Solution: Use the limit laws. \(\lim_{x\to b}[3f(x)-4g(x)]=3\lim_{x\to b}f(x)-4\lim_{x\to b}g(x)\). Substitute the given limits: \(3(2)-4(10)=6-40=-34\). Answer: \(-34\)
7.
Question 7 \(\lim_{x\to \infty} \frac{3x-2}{x^3-1}=\)
Solution: The degree of the numerator is \(1\), and the degree of the denominator is \(3\). Since the denominator grows faster than the numerator as \(x\to\infty\), the fraction approaches \(0\). Equivalently, divide by \(x^3\): \(\frac{3x-2}{x^3-1}=\frac{\frac{3}{x^2}-\frac{2}{x^3}}{1-\frac{1}{x^3}}\). As \(x\to\infty\), this becomes \(\frac{0-0}{1-0}=0\). Answer: \(0\)
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