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Question 1. The graph of a function \(f\) is shown in the figure below. Which of the following statements about \(f\) is true?
From the graph, as \(x\) approaches \(a\) from the left and from the right, the function values approach \(2\). Therefore, \(\lim_{x\to a}f(x)=2\). At \(x=b\), the graph has a jump, so the limit at \(b\) does not exist. Answer: C
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Question 2. \(\lim_{x\to -1}f(x)=?\)
From the graph, even though the function value may be different at \(x=-1\), the left-hand limit and the right-hand limit both approach \(-1\). Therefore, \(\lim_{x\to -1}f(x)=-1\). Answer: D
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Question 3. Select the best answer. \(\lim_{x\to\infty}\frac{\ln(|x|-1)}{(\cos x)^x}=?\)
As \(x\to\infty\), the numerator \(\ln(|x|-1)\) increases without bound. However, the denominator \((\cos x)^x\) oscillates because \(\cos x\) oscillates between \(-1\) and \(1\). Since the expression does not approach one fixed value, the limit does not exist. Answer: D
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Question 4. Evaluate algebraically. \(\lim_{x\to0}\frac{x}{x^2-x^2\cos x}=?\)
First factor the denominator: \(x^2-x^2\cos x=x^2(1-\cos x)\). Then \(\frac{x}{x^2-x^2\cos x}=\frac{1}{x(1-\cos x)}\). As \(x\to0^+\), \(x(1-\cos x)>0\), so the expression approaches \(\infty\). As \(x\to0^-\), \(x(1-\cos x)<0\), so the expression approaches \(-\infty\). Since the one-sided limits are different, the limit does not exist. Answer: D
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Question 5. True or False: \(\lim_{x\to -3}\frac{x^2+3x-1}{x-2}=\frac{1}{5}\).
Substitute \(x=-3\) because the denominator is not zero there. The numerator is \((-3)^2+3(-3)-1=9-9-1=-1\). The denominator is \(-3-2=-5\). Therefore, the limit is \(\frac{-1}{-5}=\frac{1}{5}\). The statement is true. Answer: A
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Question 6. What is the average rate of change of \(y\) with respect to \(x\) over the interval \([-3,-1]\) for the function \(y=5x+2\)?
The average rate of change is \(\frac{y_2-y_1}{x_2-x_1}\). For \(y=5x+2\), \(y(-3)=5(-3)+2=-13\), and \(y(-1)=5(-1)+2=-3\). Then \(\frac{-3-(-13)}{-1-(-3)}=\frac{10}{2}=5\). Answer: D
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Question 7. The Intermediate Value Theorem can be used to prove that the equation \(x^2+3x-18=0\) has at least one solution on the interval \([4,6]\).
Let \(f(x)=x^2+3x-18\). Since \(f(x)\) is a polynomial, it is continuous on \([4,6]\). Check the endpoints: \(f(4)=4^2+3(4)-18=10\), and \(f(6)=6^2+3(6)-18=36\). Both values are positive, so there is no sign change. The Intermediate Value Theorem cannot be used to prove a zero on this interval. Answer: B
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Question 8. \(\lim_{x\to0}\frac{\sin(-6x)}{\sin(-5x)}=?\)
Use \(\lim_{u\to0}\frac{\sin u}{u}=1\). Rewrite \(\frac{\sin(-6x)}{\sin(-5x)}=\frac{-6}{-5}\cdot\frac{\sin(-6x)}{-6x}\cdot\frac{-5x}{\sin(-5x)}\). As \(x\to0\), both limit factors approach \(1\). Therefore, the limit is \(\frac{-6}{-5}=\frac{6}{5}\). Answer: C
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Question 9. \(\lim_{x\to -4}\frac{6}{x+4}=?\)
As \(x\to -4^-\), \(x+4\to0^-\), so \(\frac{6}{x+4}\to-\infty\). As \(x\to -4^+\), \(x+4\to0^+\), so \(\frac{6}{x+4}\to\infty\). Since the one-sided limits are different, the two-sided limit does not exist. Answer: D
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Question 10. \(\lim_{x\to4}\frac{4-x}{\sqrt{x}-2}=?\)
Rationalize the denominator by multiplying by \(\frac{\sqrt{x}+2}{\sqrt{x}+2}\). Then \(\frac{4-x}{\sqrt{x}-2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}=\frac{(4-x)(\sqrt{x}+2)}{x-4}\). Since \(4-x=-(x-4)\), this simplifies to \(-(\sqrt{x}+2)\). Substitute \(x=4\): \(-(\sqrt{4}+2)=-(2+2)=-4\). Answer: A
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Question 11. \(\lim_{x\to\infty}\frac{4x^3+8x^2+6x}{-5x^2-9x+9}=?\)
Compare the highest-degree terms. The numerator has leading term \(4x^3\), and the denominator has leading term \(-5x^2\). Their ratio is \(\frac{4x^3}{-5x^2}=-\frac{4}{5}x\). As \(x\to\infty\), \(-\frac{4}{5}x\to-\infty\). Therefore, the limit is \(-\infty\). Answer: A
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Question 12. Find the value of \(k\) that will make the function \(f(x)\) continuous everywhere. \(f(x)=\begin{cases}6x+k, & x\le 4\\ kx^2-6, & x>4\end{cases}\)
For continuity at \(x=4\), the two pieces must have the same value at \(x=4\). Set \(6(4)+k=k(4)^2-6\). This gives \(24+k=16k-6\). Add \(6\) to both sides: \(30+k=16k\). Then \(30=15k\), so \(k=2\). Answer: C
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