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Question 1. Use the provided graph of \(f\) to evaluate \(\displaystyle \lim_{x\to 5} f(x)\).
Read the graph as \(x\) approaches \(5\) from both sides. The graph approaches the point with \(y=4\) from the left, and it also approaches \(y=4\) from the right. Since the left-hand and right-hand limits are equal, \(\displaystyle \lim_{x\to 5}f(x)=4\). Answer: D. \(4\)
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Question 2. Evaluate \(\displaystyle \lim_{x\to 0^+}\frac{\sin(2x)}{x^2}\).
Rewrite the expression as \(\displaystyle \frac{\sin(2x)}{x^2}=\frac{\sin(2x)}{2x}\cdot\frac{2}{x}\). As \(x\to 0^+\), \(\displaystyle \frac{\sin(2x)}{2x}\to 1\), while \(\displaystyle \frac{2}{x}\to \infty\). Therefore the product approaches \(\infty\). Answer: A. \(\infty\)
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Question 3. Use the provided graph of \(f\) to evaluate \(\displaystyle \lim_{x\to 7} f(x)\).
Follow the graph toward \(x=7\) from both the left and the right. Both branches approach the open point at \(y=-1\). Thus the one-sided limits agree and \(\displaystyle \lim_{x\to 7}f(x)=-1\). Answer: C. \(-1\)
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Question 4. Suppose \(\displaystyle s(n)=\sum_{i=1}^{n}\left(1+\frac{2i}{n}\right)\left(\frac{2}{n}\right)\). Find \(\displaystyle \lim_{n\to\infty}s(n)\).
This is a Riemann sum with \(\Delta x=\frac{2}{n}\), \(x_i=\frac{2i}{n}\), and \(f(x)=1+x\) on \([0,2]\). Therefore, \(\displaystyle \lim_{n\to\infty}s(n)=\int_0^2(1+x)\,dx\). Compute the integral: \(\displaystyle \left[x+\frac{x^2}{2}\right]_0^2=2+\frac{4}{2}=4\). Answer: D. \(4\)
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Question 5. Evaluate \(\displaystyle \lim_{n\to 3}\frac{\frac{1}{n}-\frac{1}{3}}{n-3}\).
Combine the fractions in the numerator: \(\displaystyle \frac{1}{n}-\frac{1}{3}=\frac{3-n}{3n}=-\frac{n-3}{3n}\). Then \(\displaystyle \frac{\frac{1}{n}-\frac{1}{3}}{n-3}=\frac{-\frac{n-3}{3n}}{n-3}=-\frac{1}{3n}\), for \(n\ne 3\). Now let \(n\to 3\): \(\displaystyle -\frac{1}{3(3)}=-\frac{1}{9}\). Answer: A. \(-\frac{1}{9}\)
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Question 6. Evaluate \(\displaystyle \lim_{x\to \pi}\frac{\sin x}{\pi-x}\).
Let \(u=\pi-x\). Then \(u\to 0\) as \(x\to\pi\), and \(\sin x=\sin(\pi-u)=\sin u\). Therefore, \(\displaystyle \frac{\sin x}{\pi-x}=\frac{\sin u}{u}\). Using \(\displaystyle \lim_{u\to 0}\frac{\sin u}{u}=1\), the limit is \(1\). Answer: A. \(1\)
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Question 7. Assume \(\displaystyle f(x)=\left(\frac{1}{x}\right)\left(\frac{1}{\sqrt{1+x}-1}\right)\). Find \(\displaystyle \lim_{x\to 0}f(x)\).
Rationalize the second denominator: \(\displaystyle \frac{1}{\sqrt{1+x}-1}\cdot\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}=\frac{\sqrt{1+x}+1}{x}\). Thus \(\displaystyle f(x)=\frac{1}{x}\cdot\frac{\sqrt{1+x}+1}{x}=\frac{\sqrt{1+x}+1}{x^2}\). As \(x\to 0\), the numerator approaches \(2\) and \(x^2\to 0^+\), so the quotient approaches \(\infty\). Answer: E. \(\infty\)
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Question 8. Suppose \(\displaystyle s(n)=\sum_{i=1}^{n}\frac{10i-n}{n^2}\). Find \(\displaystyle \lim_{n\to\infty}s(n)\).
Split the sum: \(\displaystyle s(n)=\frac{10}{n^2}\sum_{i=1}^{n}i-\frac{1}{n^2}\sum_{i=1}^{n}n\). Use \(\displaystyle \sum_{i=1}^{n}i=\frac{n(n+1)}{2}\) and \(\displaystyle \sum_{i=1}^{n}n=n^2\). Then \(\displaystyle s(n)=\frac{10}{n^2}\cdot\frac{n(n+1)}{2}-1=5\frac{n+1}{n}-1=4+\frac{5}{n}\). As \(n\to\infty\), \(\frac{5}{n}\to 0\), so the limit is \(4\). Answer: D. \(4\)
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Question 9. Evaluate \(\displaystyle \lim_{x\to 3^+}\frac{5x^2}{9-x^2}\).
Factor the denominator: \(\displaystyle 9-x^2=(3-x)(3+x)\). As \(x\to 3^+\), \(3-x\to 0^-\) and \(3+x\to 6^+\), so the denominator approaches \(0^-\). The numerator \(5x^2\to 45^+\). A positive number divided by a very small negative number approaches \(-\infty\). Answer: A. \(-\infty\)
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Question 10. Evaluate \(\displaystyle \lim_{x\to 1^-}\frac{x^2}{(1-x)(1+x)}\).
As \(x\to 1^-\), \(x^2\to 1^+\), \(1-x\to 0^+\), and \(1+x\to 2^+\). Therefore, the denominator \((1-x)(1+x)\to 0^+\). A positive numerator approaching \(1\) divided by a positive denominator approaching \(0\) grows without bound. Hence the limit is \(\infty\). Answer: E. \(\infty\)
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