Free Math Worksheets

- Divide by the highest power of \(x\) in the denominator. - Determine what happens to terms with \(x\) in the denominator as \(x\to\infty\).

1. Divide the numerator and denominator by \(x^2\): \(\frac{1+\frac{4}{x}}{2+\frac{3}{x^2}}\). 2. As \(x\to\infty\), both \(\frac{4}{x}\) and \(\frac{3}{x^2}\) approach \(0\). 3. Therefore, the limit is \(\frac{1}{2}\).

\(\frac{1}{2}\)

- Expand the numerator first. - Divide by the highest power of \(x\). - Identify which terms approach zero.

1. Expand the numerator: \((2x-1)(x+4)=2x^2+7x-4\). 2. Divide the numerator and denominator by \(x^2\): \(\frac{2+\frac{7}{x}-\frac{4}{x^2}}{5-\frac{3}{x}}\). 3. As \(x\to\infty\), the terms containing \(\frac{1}{x}\) or \(\frac{1}{x^2}\) approach \(0\). 4. Therefore, the limit is \(\frac{2}{5}\).

\(\frac{2}{5}\)

- Compare the degrees of the numerator and denominator. - Identify the dominant term in the denominator. - Decide what happens when a fixed number is divided by an increasingly large magnitude.

1. The numerator is constant, while the denominator is a third-degree polynomial. 2. As \(|x|\to\infty\), the term \(x^3\) dominates the denominator, so \(|x^3+5x|\to\infty\). 3. A constant divided by a quantity whose magnitude grows without bound approaches \(0\). Thus, \(g(x)\to0\) as \(x\to\infty\) and as \(x\to-\infty\). 4. Therefore, the horizontal asymptote is the x-axis.

\(y=0\)

- Identify the term that controls each function for large positive \(x\). - Recall the range and repeated behavior of sine. - Simplify the rational expression before taking its limit.

1. For A, the term \(-x^2\) dominates, so \(f(x)\to-\infty\). Thus, A matches (3). 2. For B, the outputs continue to oscillate between \(-1\) and \(3\), so no limit exists. Thus, B matches (4). 3. Rewrite C as \(h(x)=5+\frac{2}{x}\). Since \(\frac{2}{x}\to0\), \(h(x)\to5\). Thus, C matches (1). 4. For D, the exponent grows without bound, so \(e^{0.5x}\to+\infty\). Thus, D matches (2).

A: (3) B: (4) C: (1) D: (2)

- Track how the sign changes with the index. - Analyze the fractional factor and the sign factor separately. - Compare the behavior of the even-indexed and odd-indexed subsequences. - Recall the relationship between accumulation points and a sequence limit.

1. Substitution gives \(a_1=-2\), \(a_2=1.5\), \(a_3=-\frac43\approx-1.33\), \(a_4=1.25\), \(a_5=-1.2\), and \(a_6=\frac76\approx1.17\). 2. For even indices \(n=2k\), \(a_{2k}=\frac{2k+1}{2k}=1+\frac{1}{2k}\to1\). For odd indices \(n=2k-1\), \(a_{2k-1}=-\frac{2k}{2k-1}=-\left(1+\frac{1}{2k-1}\right)\to-1\). Therefore, the accumulation points are \(1\) and \(-1\). 3. A convergent sequence has only one accumulation point, its limit. Since this sequence has two distinct accumulation points, it diverges.

1) \(a_1=-2\), \(a_2=1.5\), \(a_3=-\frac43\), \(a_4=1.25\), \(a_5=-1.2\), and \(a_6=\frac76\). 2) The accumulation points are \(1\) and \(-1\). 3) The sequence diverges because it has two distinct accumulation points.

- Rewrite the sequence term so that the limit is easy to identify. - Express closeness to the limit with an absolute-value inequality. - Compare consecutive terms to establish monotonicity. - Use the sign of the remaining fraction to compare each term with \(2\).

1. The first six terms are \(3\), \(2.5\), \(2.33\), \(2.25\), \(2.2\), and \(2.17\), with the repeating decimals rounded to the nearest hundredth. 2. Rewrite \(a_n=2+\frac1n\). Since \(\frac1n\to0\), the limit is \(g=2\). 3. Solve \(|a_n-2|=\frac1n<0.05\). This requires \(n>20\), so the condition holds beginning at \(n=21\). 4. The difference of consecutive terms is \(a_{n+1}-a_n=\frac{1}{n+1}-\frac1n=-\frac{1}{n(n+1)}<0\). Thus, the sequence is strictly decreasing. Also, \(a_n=2+\frac1n>2\) for every positive integer \(n\).

1) \(3\), \(2.5\), \(2.33\), \(2.25\), \(2.2\), and \(2.17\). 2) \(g=2\). 3) The condition holds beginning at \(n=21\). 4) The sequence is strictly decreasing, and \(a_n>2\) for every positive integer \(n\).

- Find the horizontal asymptote. - Use half of the total band width as the maximum allowed vertical distance. - Compare \(|f(x)-2|\) with that distance.

1. The horizontal asymptote is \(y=2\). A band of total width \(0.2\) extends \(0.1\) above and below its center line, so a point is inside when \(|f(x)-2|<0.1\). 2. At \(x=15\), \(|f(15)-2|=\left|\frac{4}{10}\right|=0.4\). Since \(0.4>0.1\), the graph is outside the band. 3. At \(x=55\), \(|f(55)-2|=\left|\frac{4}{50}\right|=0.08\). Since \(0.08<0.1\), the graph is inside the band.

1. Outside 2. Inside

- Find the horizontal asymptote and half the band width. - Evaluate the vertical distance from the graph to the asymptote at \(x=10\). - Solve the distance inequality for \(x\).

1. The horizontal asymptote is \(y=5\). The band extends \(0.25\) above and below it, so points inside satisfy \(|g(x)-5|<0.25\). 2. At \(x=10\), \(|g(10)-5|=\frac{10}{21}\approx0.476\). Since this is greater than \(0.25\), the point is outside. 3. For \(x>19.5\), the denominator \(2x+1\) is positive. Solve \(\frac{10}{2x+1}<0.25\): \(10<0.25(2x+1)\), so \(40<2x+1\), which gives \(x>19.5\). 4. Thus, every graph point with \(x>19.5\) is inside the band.

1. Outside, because \(\frac{10}{21}\approx0.476>0.25\). 2. The inequality \(|g(x)-5|<0.25\) holds for every \(x>19.5\).

- Find where the denominator is zero and what value the function approaches as \(|x|\) grows. - Use half the total band width. - Compare each vertical distance from the horizontal asymptote with \(0.4\).

1. The denominator is zero at \(x=1\), so the vertical asymptote is \(x=1\). As \(|x|\to\infty\), the fraction approaches \(0\), so the horizontal asymptote is \(y=1.5\). 2. The band extends \(0.4\) above and below \(y=1.5\). A point is inside when \(|f(x)-1.5|<0.4\). 3. At \(x=3\), \(f(3)=1.5-\frac{2}{2}=0.5\). Its distance from the asymptote is \(1\), so it is outside. 4. At \(x=7\), \(f(7)=1.5-\frac{2}{6}=\frac{7}{6}\approx1.17\). Its distance from the asymptote is \(\frac{1}{3}<0.4\), so it is inside.

a) Vertical asymptote: \(x=1\); horizontal asymptote: \(y=1.5\) b) At \(x=3\): outside. At \(x=7\): inside.

- Find where the denominator is zero and what output the function approaches at infinity. - Express vertical distance from the horizontal asymptote with an absolute value. - Use the condition \(x>-2\) to determine the sign of the denominator.

1. The denominator is zero at \(0.5x+1=0\), so the vertical asymptote is \(x=-2\). As \(|x|\to\infty\), the fraction approaches \(0\), so the horizontal asymptote is \(y=-2\). 2. The required condition is \(|g(x)+2|<0.1\), or \(\left|\frac{3}{0.5x+1}\right|<0.1\). 3. For \(x>-2\), the denominator is positive, so \(\frac{3}{0.5x+1}<0.1\). 4. Multiply by the positive denominator: \(3<0.1(0.5x+1)=0.05x+0.1\). Thus, \(2.9<0.05x\), which gives \(x>58\).

a) Vertical asymptote: \(x=-2\); horizontal asymptote: \(y=-2\) b) \(x>58\)

- Determine what happens to \(3^{-x}\) as \(x\) grows. - Use half the total band width. - Compare each output's distance from \(4\) with \(0.25\).

1. Since \(3^{-x}\to0\) as \(x\to\infty\), \(f(x)\to4\). The horizontal asymptote is \(y=4\). 2. The band extends \(0.25\) above and below the asymptote, so points inside satisfy \(|f(x)-4|<0.25\). 3. At \(x=3\), \(f(3)=4+\frac{10}{27}\approx4.37\). Its distance from the asymptote is about \(0.37>0.25\), so it is outside. 4. At \(x=4\), \(f(4)=4+\frac{10}{81}\approx4.12\). Its distance from the asymptote is about \(0.12<0.25\), so it is inside.

Horizontal asymptote: \(y=4\) At \(x=3\): outside At \(x=4\): inside

- Rewrite \(2^{2-x}\) as a constant times an exponential decay factor. - Use half the total band width. - Compare the exact distances from the asymptote.

1. Since \(2^{2-x}=4\left(\frac{1}{2}\right)^x\to0\), the limit is \(5\), and the horizontal asymptote is \(y=5\). 2. The band extends \(0.05\) above and below the asymptote, so points inside satisfy \(|g(x)-5|<0.05\). 3. \(g(6)=5-2^{-4}=5-0.0625=4.9375\). The distance from \(5\) is \(0.0625>0.05\), so the point is outside. 4. \(g(8)=5-2^{-6}=5-0.015625=4.984375\). The distance from \(5\) is \(0.015625<0.05\), so the point is inside.

The limit is \(5\). At \(x=6\): outside At \(x=8\): inside

- Divide by the highest power of \(x\) in the denominator. - Evaluate the limits of the resulting reciprocal-power terms. - Treat positive and negative infinity separately, then compare the results.

1. Divide the numerator and denominator by \(x^2\): \(f(x)=\frac{\frac{4}{x}+\frac{1}{x^2}}{2-\frac{3}{x^2}}\). 2. As \(x\to\infty\) or \(x\to-\infty\), each term containing \(\frac{1}{x}\) or \(\frac{1}{x^2}\) approaches \(0\). 3. Therefore, the numerator approaches \(0\), the denominator approaches \(2\), and \(f(x)\to0\) in both directions.

\(\lim_{x\to\infty}f(x)=0\) and \(\lim_{x\to-\infty}f(x)=0\)

- Shift a sine or cosine function to obtain the required oscillation interval. - A polynomial of degree greater than \(1\) can outgrow every linear function. - Reflect a positive growth function across the x-axis to obtain behavior toward \(-\infty\).

1. For part a, \(f(x)=\sin(x)+3\) is trigonometric and has range \([2, 4]\), so it continues to oscillate without a limit. 2. For part b, \(g(x)=x^2\) is polynomial and approaches \(+\infty\) faster than any linear function. 3. For part c, \(h(x)=-e^x\) is exponential and approaches \(-\infty\).

Possible answers: a) \(f(x)=\sin(x)+3\) b) \(g(x)=x^2\) c) \(h(x)=-e^x\)

- Use half the total band width. - Express the distance from the horizontal asymptote with an absolute value. - Solve the resulting absolute-value inequality.

1. Since \(\frac{5}{x-3}\to0\) as \(x\to\pm\infty\), the horizontal asymptote is \(y=1\). 2. The band extends \(0.1\) above and below the asymptote, so the condition is \(\left|\frac{5}{x-3}\right|<0.1\). 3. This is equivalent to \(|x-3|>50\), giving \(x>53\) or \(x<-47\). 4. Therefore, one suitable pair of thresholds is \(x_0=53\) and \(x_1=-47\).

a) \(y=1\) b) \(x_0=53\) c) \(x_1=-47\)

- Use half the total band width as the allowed deviation. - Write an inequality involving \(\frac{a}{x}\). - Solve for \(x\) while using \(a>0\) and \(x>0\).

1. The band extends \(0.005\) above and below \(y=2\). For \(x>0\), the condition is \(g_a(x)-2<0.005\). 2. Substitute the formula: \(\frac{a}{x}<0.005\). 3. Since \(a>0\) and \(x>0\), this is equivalent to \(x>\frac{a}{0.005}=200a\). 4. Thus, \(x_{\text{start}}=200a\). The threshold is directly proportional to \(a\): doubling \(a\) doubles the required threshold.

\(x_{\text{start}}=200a\); the threshold is directly proportional to \(a\).

- For a rational function with equal numerator and denominator degrees, use the ratio of leading coefficients. - Shift an exponential decay function downward. - Set each function equal to zero to find its zeros.

1. One rational example is \(f(x)=\frac{-2x}{x+1}\). The ratio of leading coefficients gives \(f(x)\to-2\) as \(x\to\infty\) and as \(x\to-\infty\). Its zero is \(x=0\). 2. One exponential example is \(g(x)=e^{-x}-2\). As \(x\to\infty\), \(e^{-x}\to0\), so \(g(x)\to-2\). As \(x\to-\infty\), \(e^{-x}\to+\infty\), so \(g(x)\to+\infty\). 3. Solve \(e^{-x}-2=0\): \(e^{-x}=2\), so \(-x=\ln(2)\) and \(x=-\ln(2)\).

Possible examples: \(f(x)=\frac{-2x}{x+1}\): as \(x\to-\infty\), \(f(x)\to-2\); zero at \(x=0\). \(g(x)=e^{-x}-2\): as \(x\to-\infty\), \(g(x)\to+\infty\); zero at \(x=-\ln(2)\).

- Distinguish between a lower bound and the actual limit. - Try a decreasing reciprocal function shifted upward. - One counterexample is enough to disprove a universal statement.

1. The statement is false. Being decreasing and bounded below by \(2\) does not force the limit to equal that particular lower bound. 2. Consider \(f(x)=3+\frac{1}{x}\) for \(x>0\). This function is decreasing and satisfies \(f(x)>3>2\). 3. However, \(\lim_{x\to\infty}\left(3+\frac{1}{x}\right)=3\), not \(2\). This counterexample disproves the statement.

False. A counterexample is \(f(x)=3+\frac{1}{x}\), which decreases and stays above \(2\) but approaches \(3\).

- Divide by \(x\) to find the limit. - Use the positive denominator to clear the inequality safely. - Distinguish an arbitrary upper bound from the value approached by the function.

1. Divide the numerator and denominator by \(x\): \(f(x)=\frac{4-\frac{1}{x}}{1+\frac{1}{x}}\). Therefore, \(L=4\). 2. Since \(x+1>0\), \(\frac{4x-1}{x+1}<5\) is equivalent to \(4x-1<5x+5\), or \(x>-6\). This is true for every \(x>0\). 3. The function can also be written as \(f(x)=4-\frac{5}{x+1}\), which increases toward \(4\). Although \(5\) is an upper bound, it is not the limit. An upper bound need not be the least upper bound or the value approached by the function.

1. \(L=4\) 2. \(f(x)<5\) for every \(x>0\). 3. The statement is false because the actual limit is \(4\), even though \(5\) is an upper bound.

- Divide the numerator and denominator by the highest power of \(n\). - Express the distance from a sequence term to the limit with an absolute value. - Solve the resulting inequality for \(n\). - To study monotonicity, compare consecutive terms.

1. Substitute the indices: \(a_1=\frac35=0.6\), \(a_2=\frac77=1\), and \(a_{10}=\frac{39}{23}\approx1.696\). 2. Divide the numerator and denominator by \(n\): \(a_n=\frac{4-\frac1n}{2+\frac3n}\to\frac42=2\). Thus, \(g=2\). 3. Solve the error inequality: \(\left|\frac{4n-1}{2n+3}-2\right|=\frac{7}{2n+3}<0.001\). This gives \(7000<2n+3\), so \(n>3498.5\). Therefore, the least possible value is \(n_0=3499\). 4. Compute the difference of consecutive terms: \(a_{n+1}-a_n=\frac{4n+3}{2n+5}-\frac{4n-1}{2n+3}=\frac{14}{(2n+5)(2n+3)}>0\). Therefore, the sequence is strictly increasing.

1) \(a_1=0.6\), \(a_2=1\), and \(a_{10}\approx1.696\). 2) \(g=2\). 3) \(n_0=3499\). 4) Since \(a_{n+1}-a_n>0\), the sequence is strictly increasing.

- Compare the degrees of the numerator and denominator. - You may analyze the derivative of the associated real-valued function or compare consecutive terms. - Rewrite the condition as a quadratic inequality. - Find the zeros of the quadratic expression.

1. Polynomial division gives \(b_n=n-5+\frac{15}{n+5}\). The linear term grows without bound while the fraction approaches \(0\), so \(\lim_{n\to\infty}b_n=\infty\). 2. Compare consecutive terms: \(b_{n+1}-b_n=\frac{(n+1)^2-10}{n+6}-\frac{n^2-10}{n+5}=\frac{n^2+11n+15}{(n+5)(n+6)}\). For every positive integer \(n\), both the numerator and denominator are positive. Therefore, the sequence is strictly increasing for all positive integers \(n\). 3. Solve \(\frac{n^2-10}{n+5}>50\). Since \(n+5>0\), this is equivalent to \(n^2-50n-260>0\). The positive root is \(25+\sqrt{885}\approx54.75\), so the least positive integer solution is \(n=55\).

1) \(\lim_{n\to\infty}b_n=\infty\). 2) Since \(b_{n+1}-b_n>0\) for every positive integer \(n\), the sequence is strictly increasing. 3) The least value is \(n=55\).

- Determine what happens to the fractional term as \(n\) grows. - Rewrite the interval condition as \(|a_n-g|<\varepsilon\). - Review the definition of a monotonic sequence. - To show boundedness, find fixed lower and upper bounds for all terms.

1. Substitution gives \(a_1=-6\), \(a_2=9\), \(a_{10}=5\), and \(a_{100}=4.1\). 2. Since \(\frac{10(-1)^n}{n}\to0\), the sequence converges to \(g=4\). 3. The interval condition is equivalent to \(|a_n-4|<0.05\). Thus, \(\left|\frac{10(-1)^n}{n}\right|=\frac{10}{n}<0.05\). This requires \(n>200\), so every term beginning with \(a_{201}\) lies in the interval. 4. The odd-indexed terms lie below \(4\), and the even-indexed terms lie above \(4\), so the sequence is not monotonic. Also, the largest deviation occurs among the first terms: \(-6\le a_n\le9\). Therefore, it is bounded.

1) \(a_1=-6\), \(a_2=9\), \(a_{10}=5\), and \(a_{100}=4.1\). 2) \(g=4\). 3) Every remaining term lies in the interval beginning at \(n=201\). 4) The sequence is bounded by \(-6\) and \(9\), but it alternates around its limit and is not monotonic.

- Split the fraction so that \(n\) appears in only one term. - Use the rewritten form to compute the distance from the limit. - Check the smallest allowed index and how the fractional term changes as \(n\) increases.

1. Rewrite the term: \(b_n=\frac{n^2}{2n^2}-\frac{1}{2n^2}=\frac12-\frac{1}{2n^2}=0.5-\frac{1}{2n^2}\). Since \(\frac{1}{2n^2}\to0\), \(b_n\to0.5\). 2. Solve \(|b_n-0.5|=\frac{1}{2n^2}<0.01\). This gives \(n^2>50\), so \(n>\sqrt{50}\approx7.071\). Therefore, the condition holds for every integer \(n\ge8\). 3. For every \(n\ge1\), \(\frac{1}{2n^2}>0\), so \(b_n<0.5\). Also, \(b_1=0\), and \(\frac{1}{2n^2}\le\frac12\), so \(b_n\ge0\). Thus, \(0\le b_n<0.5\).

1) \(b_n=0.5-\frac{1}{2n^2}\), so the limit is \(0.5\). 2) The condition holds for every \(n\ge8\). 3) The sequence satisfies \(0\le b_n<0.5\).

- Compare \(b_n\) and \(b_{n+1}\). - An upper bound is a number no sequence term exceeds. - Divide by the highest power of \(n\) to find the limit. - Solve the final inequality for \(n\), paying attention to the strict inequality.

1. Compare consecutive terms: \(b_{n+1}-b_n=\frac{3n+2}{n+2}-\frac{3n-1}{n+1}=\frac{4}{(n+1)(n+2)}>0\). Therefore, the sequence is strictly increasing. 2. Since \(b_n=\frac{3n-1}{n+1}<\frac{3n+3}{n+1}=3\), the number \(3\) is an upper bound. 3. Dividing the numerator and denominator by \(n\) gives \(L=\lim_{n\to\infty}\frac{3-\frac1n}{1+\frac1n}=3\). 4. The error is \(|3-b_n|=\frac{4}{n+1}\). The inequality \(\frac{4}{n+1}<\varepsilon\) is equivalent to \(n>\frac{4}{\varepsilon}-1\). Therefore, the least positive integer that works is \(n_0=\max\left(1, \left\lfloor\frac{4}{\varepsilon}-1\right\rfloor+1\right)\).

1) \(b_{n+1}-b_n=\frac{4}{(n+1)(n+2)}>0\), so the sequence is strictly increasing. 2) An upper bound is \(3\). 3) \(L=3\). 4) \(n_0=\max\left(1, \left\lfloor\frac{4}{\varepsilon}-1\right\rfloor+1\right)\).

- Compare the expressions under the square roots for consecutive indices. - Square the inequality carefully when both sides are positive. - Solve the resulting quadratic inequality for \(n\). - A simpler lower comparison may also establish unbounded growth.

1. Since \(a_{n+1}^2-a_n^2=[(n+1)^2+4(n+1)]-(n^2+4n)=2n+5>0\), and both terms are nonnegative, \(a_{n+1}>a_n\). Thus, the sequence is strictly increasing. 2. The inequality \(a_n>M\) is equivalent to \(n^2+4n>M^2\), or \((n+2)^2>M^2+4\). Since \(n+2>0\), it is enough to choose \(n>-2+\sqrt{M^2+4}\). Such an integer exists for every \(M>0\), so the sequence is unbounded above. 3. For \(M=100\), the condition becomes \(n>-2+\sqrt{10{,}004}\approx98.02\). Therefore, the least positive integer is \(n=99\). Since the sequence is increasing, every later term is also greater than \(100\).

1) The sequence is strictly increasing. 2) It is enough to choose \(n>-2+\sqrt{M^2+4}\). 3) The least index is \(n=99\).

- Compare consecutive terms. - Translate unbounded decrease into a condition involving an arbitrary positive number \(M\). - Set up an inequality for each requested threshold. - Remember that the index must be a positive integer.

1. Compare consecutive terms: \(a_{n+1}-a_n=-\frac14[(n+1)^2-n^2]=-\frac12n-\frac14<0\). Therefore, the sequence is strictly decreasing. 2. Let \(M>0\). The condition \(a_n<-M\) is equivalent to \(15-\frac14n^2<-M\), so \(n^2>4(15+M)\). Thus, any integer satisfying \(n>2\sqrt{15+M}\) makes \(a_n<-M\). Therefore, the sequence is unbounded below and \(a_n\to-\infty\). 3. Solve \(15-\frac14n^2<-200\). This gives \(n^2>860\), so \(n>\sqrt{860}\approx29.33\). The least index is \(n=30\). 4. Solve \(15-\frac14n^2<-5000\). This gives \(n^2>20{,}060\), so \(n>\sqrt{20{,}060}\approx141.63\). The least index is \(n=142\).

1) The sequence is strictly decreasing. 2) For every \(M>0\), \(a_n<-M\) whenever \(n>2\sqrt{15+M}\), so \(a_n\to-\infty\). 3) Beginning at \(n=30\). 4) Beginning at \(n=142\).

- Use trigonometry in one isosceles triangular section to find the side length. - Apply the double-angle identity for sine in part b). - Determine what happens to the cosine argument as \(n\) grows.

1. The central angle subtended by one side is \(\frac{2\pi}{n}\). Bisecting the corresponding isosceles triangle gives side length \(s_n=2R\sin\left(\frac{\pi}{n}\right)\). Therefore, \(p_n=2nR\sin\left(\frac{\pi}{n}\right)\). 2. Replacing \(n\) by \(2n\), \(p_{2n}=4nR\sin\left(\frac{\pi}{2n}\right)\). Thus, \(v_n=\frac{2\sin(\pi/(2n))}{\sin(\pi/n)}\). 3. Using \(\sin(2x)=2\sin x\cos x\) with \(x=\frac{\pi}{2n}\), \(\sin\left(\frac{\pi}{n}\right)=2\sin\left(\frac{\pi}{2n}\right)\cos\left(\frac{\pi}{2n}\right)\). Therefore, \(v_n=\frac{1}{\cos\left(\frac{\pi}{2n}\right)}\). 4. Since \(\frac{\pi}{2n}\to0\), \(\lim_{n\to\infty}v_n=\frac{1}{\cos0}=1\). The relative increase \(\frac{p_{2n}-p_n}{p_n}=v_n-1\) approaches \(0\), so doubling the number of sides produces a vanishing relative improvement as the polygonal perimeter approaches the circle's circumference.

a) \(p_n=2nR\sin\left(\frac{\pi}{n}\right)\). b) \(v_n=\frac{1}{\cos\left(\frac{\pi}{2n}\right)}\). c) The limit is \(1\); the relative perimeter increase from doubling the number of sides approaches \(0\).

- Consider what the graph might do after it leaves the right edge of the window. - A decreasing curve does not necessarily continue decreasing forever. - Distinguish visual evidence from a mathematical proof of a limit.

1. In the visible interval, the graph decreases and moves closer to the x-axis. 2. A finite graphing window does not determine what happens for all larger x-values. The graph could later turn upward, continue downward, or approach a different horizontal asymptote. 3. For example, \(p(x)=\frac{10}{x+1}+0.02x\) decreases throughout the displayed interval but eventually rises without bound because the linear term dominates. Thus, a graph that looks as though it approaches \(0\) in one window need not have limit \(0\). 4. Therefore, Lukas’s conclusion is not justified without additional information, such as a function rule or a proof.

No. A finite graphing window can suggest a limit but cannot establish the behavior as \(x\to\infty\). The graph may behave differently outside the displayed interval.

- Use two convenient points on the dashed line to find its slope. - Think about how the secant line changes as point \(P\) moves along the curve toward \(O\). - The expression in the limit is a difference quotient at \(x=0\).

1. The secant slope is \(m_s = \frac{f(3)-f(0)}{3-0} = \frac{1-0}{3} = \frac{1}{3}\). 2. The tangent line passes through \((0, 0)\) and \((2, 1)\), so its slope is \(m_t = \frac{1-0}{2-0} = 0.5\). 3. The expression \(\frac{\sqrt{x+1}-1}{x}\) is the slope of the secant line through \(O\) and \((x,f(x))\). As \(x\to 0\), the secant line approaches the tangent line at the origin. 4. Therefore, the secant slopes approach the tangent slope, so \(\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = 0.5\).

a) \(m_s = \frac{1}{3}\) b) \(m_t = 0.5\) c) The limit is the tangent slope at the origin, so \(\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = 0.5\).

- Compare \(a_{n+1}\) and \(a_n\) by subtracting. - Divide the numerator and denominator by the highest power of \(n\). - Solve \(|a_n-g|<\varepsilon\) for \(n\). - Use monotonicity and the limit to identify bounds.

1. Compare consecutive terms: \(a_{n+1}-a_n=\frac{3(n+1)^2}{(n+1)^2+2}-\frac{3n^2}{n^2+2}\) \(=\frac{12n+6}{((n+1)^2+2)(n^2+2)}>0\). Therefore, the sequence is strictly increasing. 2. Divide the numerator and denominator by \(n^2\): \(a_n=\frac{3}{1+\frac{2}{n^2}}\to3\). Thus, \(g=3\). 3. Solve \(\left|\frac{3n^2}{n^2+2}-3\right|=\frac{6}{n^2+2}<0.001\). This requires \(n^2>5998\), or \(n>\sqrt{5998}\approx77.45\). Since the condition must hold for every \(n>N\), the least value is \(N=77\). 4. The first term is \(a_1=1\). Because the sequence is strictly increasing and converges to \(3\), \(1\) is a lower bound and \(3\) is an upper bound.

1) \(a_{n+1}-a_n>0\), so the sequence is strictly increasing. 2) \(g=3\). 3) \(N=77\), so the condition holds for every \(n\ge78\). 4) A lower bound is \(1\), and an upper bound is \(3\).

- Divide each term in the numerator by the denominator. - Identify which part controls the behavior for large \(n\). - Convert each rational inequality into a quadratic inequality. - Check the least integer above the positive zero.

1. Rewrite \(b_n=3n-\frac{1000}{n}\). As \(n\to\infty\), the first term tends to \(\infty\) and the second tends to \(0\). Therefore, \(\lim_{n\to\infty}b_n=\infty\). 2. Solve \(3n-\frac{1000}{n}>100\). Since \(n>0\), this is equivalent to \(3n^2-100n-1000>0\). The positive zero is \(\frac{100+\sqrt{22{,}000}}{6}\approx41.39\). Thus, the least possible value is \(n_0=42\). 3. Similarly, \(b_n>1000\) is equivalent to \(3n^2-1000n-1000>0\). The positive zero is \(\frac{1000+\sqrt{1{,}012{,}000}}{6}\approx334.33\). Therefore, \(n_1=335\).

1) \(\lim_{n\to\infty}b_n=\infty\). 2) \(n_0=42\). 3) \(n_1=335\).

- Use the slope formula between \(O\) and \(P\). - Read the tangent slope from two grid points. - Imagine moving \(P\) closer to the origin. - Connect the limiting secant slope to the tangent slope.

1. The secant slope is \(m_s = \frac{\sin(x)-\sin(0)}{x-0} = \frac{\sin(x)}{x}\). 2. The tangent line passes through \((0, 0)\) and \((1, 1)\), so its slope is \(1\). 3. As \(x\to 0\), point \(P\) approaches \(O\), and the secant line approaches the tangent line. 4. Therefore, the secant slopes approach the tangent slope, so \(\lim_{x\to 0}\frac{\sin(x)}{x} = 1\).

1. \(m_s = \frac{\sin(x)}{x}\) 2. \(1\) 3. The secant approaches the tangent, and \(\lim_{x\to 0}\frac{\sin(x)}{x} = 1\).