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Question 1 The figure above shows the graph of function \(f\) with domain \(0\le x\le 4\). Which of the following statements are true? I. \(\lim_{x\to 2^-}f(x)\) exists II. \(\lim_{x\to 2^+}f(x)\) exists III. \(\lim_{x\to 2}f(x)\) exists
Solution: From the graph, the left-hand limit \(\lim_{x\to 2^-}f(x)\) exists and the right-hand limit \(\lim_{x\to 2^+}f(x)\) exists. However, the two one-sided limits are not equal, so the two-sided limit \(\lim_{x\to 2}f(x)\) does not exist. Therefore, statements I and II are true, but statement III is false. Answer: I and II only
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Question 2 Evaluate \(\lim_{x\to 3} f(x)\).
Solution: From the graph, there is a vertical asymptote at \(x=3\). As \(x\to3^-\), \(f(x)\to-\infty\), and as \(x\to3^+\), \(f(x)\to-\infty\). Since both one-sided limits approach \(-\infty\), the limit is \(-\infty\). Answer: \(-\infty\)
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Question 3 Select the best answer. \(\lim_{x\to\infty}(3^x+2x)(\sin x)^{x+1}\).
Solution: The factor \(3^x+2x\) grows without bound as \(x\to\infty\). However, \((\sin x)^{x+1}\) does not approach a single stable value because \(\sin x\) oscillates between \(-1\) and \(1\). Therefore the whole expression does not approach one fixed limit. Answer: DNE
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Question 4 Evaluate the following algebraically. \(\lim_{x\to0}\frac{\csc^2 x}{\frac{1}{x^2}}\).
Solution: Simplify the expression: \(\frac{\csc^2 x}{\frac{1}{x^2}}=x^2\csc^2x=\frac{x^2}{\sin^2x}=\left(\frac{x}{\sin x}\right)^2\). Since \(\lim_{x\to0}\frac{\sin x}{x}=1\), we also have \(\lim_{x\to0}\frac{x}{\sin x}=1\). Therefore the limit is \(1^2=1\). Answer: 1
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Question 5 Find \(\lim_{x\to3}\frac{x^2+4x-1}{x-6}\).
Solution: Substitute \(x=3\) because the denominator is not zero at \(x=3\). We get \(\frac{3^2+4(3)-1}{3-6}=\frac{9+12-1}{-3}=\frac{20}{-3}=-\frac{20}{3}\). Answer: \(-\frac{20}{3}\)
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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 \(y=5x+2\)?
Solution: 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 \(y(-1)=5(-1)+2=-3\). Thus the average rate of change is \(\frac{-3-(-13)}{-1-(-3)}=\frac{10}{2}=5\). Answer: 5
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Question 7 The Intermediate Value Theorem can be used to prove that the equation \(x^2-2x-15=0\) has at least one solution on the interval \([4,6]\).
Solution: Let \(f(x)=x^2-2x-15\). This is a polynomial, so it is continuous on \([4,6]\). Evaluate the endpoints: \(f(4)=4^2-2(4)-15=16-8-15=-7\), and \(f(6)=6^2-2(6)-15=36-12-15=9\). Since \(f(4)0\), the function changes sign. By the Intermediate Value Theorem, there is at least one solution on \([4,6]\). Answer: Ture
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Question 8 Evaluate \(\lim_{x\to0}\frac{\sin(-2x)}{\sin(-4x)}\).
Solution: Use the standard sine limit. Rewrite \(\frac{\sin(-2x)}{\sin(-4x)}=\frac{-2}{-4}\cdot\frac{\sin(-2x)}{-2x}\cdot\frac{-4x}{\sin(-4x)}\). As \(x\to0\), both sine limit factors approach 1. Therefore the limit is \(\frac{-2}{-4}\cdot1\cdot1=\frac{1}{2}\). Answer: \(\frac{1}{2}\)
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Question 9 What is \(\lim_{x\to -2^-}\frac{x}{x+2}\)?
Solution: As \(x\to-2^-\), the numerator \(x\to-2\), which is negative, and the denominator \(x+2\to0^-\), which is also negative. A negative number divided by a very small negative number becomes positive without bound. Therefore the limit is \(\infty\). Answer: \(\infty\)
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Question 10 Evaluate \(\lim_{x\to9}\frac{9-x}{\sqrt{x}-3}\).
Solution: Rationalize the denominator by multiplying by \(\frac{\sqrt{x}+3}{\sqrt{x}+3}\). Then \(\frac{9-x}{\sqrt{x}-3}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{(9-x)(\sqrt{x}+3)}{x-9}=-(\sqrt{x}+3)\). Now substitute \(x=9\): \(-(\sqrt{9}+3)=-(3+3)=-6\). Answer: -6
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Question 11 Evaluate \(\lim_{x\to\infty}\frac{-4x^3+5x^2-7x}{-7x^2+3x+9}\).
Solution: Compare the leading terms. The numerator has degree 3 and the denominator has degree 2. For large \(x\), the expression behaves like \(\frac{-4x^3}{-7x^2}=\frac{4}{7}x\). As \(x\to\infty\), \(\frac{4}{7}x\to\infty\). Answer: \(\infty\)
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Question 12 Find the value of \(k\) that will make the function \(f(x)\) continuous everywhere. \(f(x)=\begin{cases}2x+k, & x\le -2\\ kx^2-7, & x>-2\end{cases}\).
Solution: For continuity at \(x=-2\), the two pieces must be equal there. Set \(2(-2)+k=k(-2)^2-7\). This gives \(-4+k=4k-7\). Add 7 to both sides: \(3+k=4k\). Thus \(3=3k\), so \(k=1\). Answer: 1
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