Free math worksheets

- Compare values of \(f(x)\) for inputs just below and just above \(3\). - Focus on the number the outputs approach, not on whether \(x=3\) appears in the table. - Check that the values from both sides suggest the same target.

1. From the left of \(3\), the values \(6.8\) and \(6.98\) are getting close to \(7\). 2. From the right of \(3\), the values \(7.02\) and \(7.2\) also approach \(7\) as \(x\) gets closer to \(3\). 3. Therefore, the table supports the estimate \(\lim_{x\to 3} f(x)=7\).

\(\lim_{x\to 3} f(x)=7\)

- Think about how the graph of a linear function behaves near the given x-value. - The approaching value is the y-value the line tends toward near that input. - Check the function value at the input after considering the nearby behavior.

1. A linear function has no break at \(x=-2\), so its nearby values approach the function value there. 2. Compute \(f(-2)=3(-2)+7=1\). 3. Therefore, \(\lim_{x\to -2}f(x)=1\).

\(\lim_{x\to -2}f(x)=1\)

- Look at the graph just to the left and just to the right of \(x=1\). - Focus on the \(y\)-value the graph approaches, not on points far from \(x=1\).

1. Follow the graph toward \(x=1\) from values on both sides. 2. The \(y\)-values approach \(3\), so \(\lim_{x\to 1}f(x)=3\).

\(3\)

- Check whether the expression is defined at \(x=-1\). - For a polynomial at an ordinary input, compare the limit with the value obtained by substitution.

1. The polynomial is continuous at \(x=-1\), so substitute \(-1\). 2. \((-1)^2+3(-1)+4=1-3+4=2\). 3. Therefore, the limit is \(2\).

\(2\)

- 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)

- Try rewriting the numerator using a multiple of the denominator. - In the rewritten form, identify the part that becomes small as \(n\) grows. - Check the sign of the term being added to or subtracted from the limiting value.

1. Substitution gives \(a_1=\frac35=0.6\), \(a_2=1\), and \(a_{10}=\frac{39}{23}\approx1.696\). 2. Since \(4n-1=2(2n+3)-7\), \(a_n=2-\frac{7}{2n+3}\). As \(n\) increases, \(\frac{7}{2n+3}\) approaches \(0\), so \(a_n\) approaches \(2\). 3. The subtracted fraction is positive for every positive integer \(n\), so every term is below \(2\). The terms therefore approach \(2\) from below.

1) \(a_1=0.6\), \(a_2=1\), and \(a_{10}\approx1.696\). 2) \(a_n=2-\frac{7}{2n+3}\), so the sequence approaches \(2\). 3) The terms approach \(2\) from below.

- Divide the quadratic expression by the linear denominator. - Compare how the two terms in the rewritten expression behave for large \(n\). - Decide whether either term stays bounded while the other grows.

1. Polynomial division gives \(b_n=n-5+\frac{15}{n+5}\). 2. As \(n\to\infty\), the term \(n-5\) grows without bound while \(\frac{15}{n+5}\) approaches \(0\). Therefore, \(\lim_{n\to\infty}b_n=+\infty\). 3. The linear term \(n-5\) controls the long-run behavior because it grows without bound, while the fractional term becomes negligible.

1) \(b_n=n-5+\frac{15}{n+5}\). 2) \(\lim_{n\to\infty}b_n=+\infty\). 3) The linear term \(n-5\) controls the long-run behavior.

- Separate the constant part of the sequence from the changing fraction. - Think about what \((-1)^n\) does to the sign from one term to the next. - Focus on the size of \(\frac{10}{n}\) as \(n\) grows.

1. Substitution gives \(a_1=-6\), \(a_2=9\), \(a_{10}=5\), and \(a_{100}=4.1\). 2. The magnitude of \(\frac{10(-1)^n}{n}\) is \(\frac{10}{n}\), which approaches \(0\). Therefore, \(a_n\) approaches \(4\). 3. The factor \((-1)^n\) changes sign from one term to the next, so the terms alternate above and below \(4\). At the same time, \(\frac{10}{n}\) decreases toward \(0\), so the distance from \(4\) shrinks.

1) \(a_1=-6\), \(a_2=9\), \(a_{10}=5\), and \(a_{100}=4.1\). 2) The sequence approaches \(4\). 3) \((-1)^n\) alternates the sign, while \(\frac{10}{n}\) makes the distance from \(4\) shrink.

- Try expressing the numerator as a multiple of the denominator plus a constant remainder. - In the rewritten form, identify the part that becomes small as \(n\) grows. - The sign of the remaining fraction tells you which side of the limiting value the terms lie on.

1. \(a_1=1\), \(a_2=2\), and \(a_{10}=\frac{300}{102}=\frac{50}{17}\approx2.941\). 2. Since \(3n^2=3(n^2+2)-6\), \(a_n=3-\frac{6}{n^2+2}\). 3. The fraction \(\frac{6}{n^2+2}\) is positive and approaches \(0\), so \(a_n\to3\) from below.

1) \(a_1=1\), \(a_2=2\), and \(a_{10}=\frac{50}{17}\approx2.941\). 2) \(a_n=3-\frac{6}{n^2+2}\). 3) The sequence approaches \(3\) from below.

- 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. - Ask whether the whole sequence can have one limit if those two subsequences approach different values.

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 \(n=2k-1\), \(a_{2k-1}=-\frac{2k}{2k-1}=-\left(1+\frac{1}{2k-1}\right)\to-1\). Thus, the even-indexed terms approach \(1\), while the odd-indexed terms approach \(-1\). 3. The full sequence cannot approach one value because its even-indexed and odd-indexed terms approach two different values. Therefore, the sequence 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 even-indexed terms approach \(1\), and the odd-indexed terms approach \(-1\). 3) The sequence diverges because the two subsequences approach different values, so the full sequence has no single limit.

- Split the fraction into two simpler fractions. - Identify the part that becomes small as \(n\) increases. - Compare each selected term with the limiting value using an absolute difference.

1. Rewrite \(b_n=\frac12-\frac{1}{2n^2}\). 2. Since \(\frac{1}{2n^2}\to0\), the sequence approaches \(\frac12\). 3. \(b_2=\frac38\), so its distance from \(\frac12\) is \(\frac18\). Also, \(b_{10}=\frac{99}{200}\), so its distance from \(\frac12\) is \(\frac{1}{200}\). The later term is much closer to the limiting value.

1) \(b_n=\frac12-\frac{1}{2n^2}\). 2) The sequence approaches \(\frac12\). 3) The distances are \(\frac18\) and \(\frac{1}{200}\); the later term is much closer to \(\frac12\).

- Divide each term in the numerator by \(n\). - Focus on what happens to \(\frac1n\) for large \(n\). - Interpret \(\frac1n\) as the distance from the limiting value.

1. The first six terms are \(3\), \(2.5\), \(2.33\), \(2.25\), \(2.2\), and \(2.17\). 2. Rewrite \(a_n=2+\frac1n\). Since \(\frac1n\to0\), the sequence approaches \(2\). 3. The distance from \(a_n\) to \(2\) is \(\frac1n\), and that distance becomes smaller as \(n\) increases.

1) \(3\), \(2.5\), \(2.33\), \(2.25\), \(2.2\), and \(2.17\). 2) \(a_n=2+\frac1n\), so the sequence approaches \(2\). 3) The distance from \(2\) is \(\frac1n\), which decreases as \(n\) increases.

- Divide both terms in the numerator by \(n\). - Compare what happens to the linear term and the reciprocal term as \(n\) becomes large. - The sample terms can help you check whether your long-run description is reasonable.

1. Divide each term in the numerator by \(n\): \(b_n=3n-\frac{1000}{n}\). 2. \(b_{10}=-70\), \(b_{100}=290\), and \(b_{1000}=2999\). 3. As \(n\to\infty\), \(3n\to\infty\) while \(\frac{1000}{n}\to0\). Therefore, \(b_n\to+\infty\), and the term \(3n\) controls the long-run behavior.

1) \(b_n=3n-\frac{1000}{n}\). 2) \(b_{10}=-70\), \(b_{100}=290\), and \(b_{1000}=2999\). 3) \(b_n\to+\infty\); the term \(3n\) controls the long-run behavior.

- Follow the graph from both sides of \(x=2\) and identify the height it approaches. - Use the filled point to determine the actual function value. - Keep the behavior near an input separate from the value at that input.

1. From both sides of \(x=2\), the graph approaches the open point at height \(3\). Therefore, \(\lim_{x\to2}f(x)=3\). 2. The filled point at \(x=2\) has height \(5\), so \(f(2)=5\). 3. A limit describes nearby function values. Changing the value at the single input \(x=2\) does not change what nearby values approach.

a) \(3\) b) \(5\) c) The nearby values approach \(3\), while the defined function value is \(5\).

- Substitute \(x=3\) first and notice what form appears. - Look for a factorization of the numerator that uses the same factor as the denominator. - A limit depends on values near \(x=3\), so simplifying an equivalent expression for \(x\ne 3\) can help.

1. Direct substitution gives \(\frac{0}{0}\), which is undefined and does not determine the limit. 2. Factor the numerator: \(x^2-9=(x-3)(x+3)\). 3. For \(x\ne 3\), \(\frac{(x-3)(x+3)}{x-3}=x+3\). 4. The simplified expression approaches \(3+3=6\) as \(x\to 3\), so the limit is \(6\).

\(6\)

- Separate the question “What is the function value?” from “What do nearby values approach?” - For inputs close to \(1\) but not equal to \(1\), which rule applies? - Evaluate the nearby rule as its input approaches \(1\).

1. The statement confuses the value at \(x=1\) with the behavior of the function near \(x=1\). 2. For every nearby input other than \(1\), the function follows \(2x+1\). 3. As \(x\) approaches \(1\), \(2x+1\) approaches \(3\). 4. Therefore, \(\lim_{x\to1}f(x)=3\), even though \(f(1)=7\).

Otieno incorrectly assumes that a function value determines the limit. Here \(\lim_{x\to1}f(x)=3\), while \(f(1)=7\).

- For the limit, first ignore the row with \(x=2\) and look at nearby rows on both sides. - Treat the exact function value and the approached value as separate questions. - Check whether the left-side and right-side values are heading toward the same number.

1. The values with \(x<2\) approach \(5\), and the values with \(x>2\) also approach \(5\). 2. Therefore, \(\lim_{x\to 2}f(x)=5\). 3. The row at \(x=2\) gives \(f(2)=9\). 4. A limit describes nearby function values as \(x\) approaches the input; it does not have to equal the function's defined value at that input.

a) \(5\) b) \(9\) c) The limit is determined by nearby values, while \(f(2)\) is the function's value at the single input \(x=2\).

- 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.

- 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 rise over run for the secant through \(O\) and \(P\). - Read two convenient points on the dashed tangent line. - Compare the successive secants with the tangent as their second endpoints move closer to \(O\).

1. The secant through \(O\) and \(P\) has slope \(\frac{1-0}{3-0}=\frac13\). 2. The tangent line passes through \((0,0)\) and \((2,1)\), so its slope is \(\frac12\). 3. The expression \(\frac{\sqrt{x+1}-1}{x}\) is the slope of the secant through \(O\) and \((x,f(x))\). The displayed secants move toward the tangent as \(x\) approaches \(0\), so the secant slopes approach \(\frac12\). Therefore, the limit is \(\frac12\).

a) \(\frac13\) b) \(\frac12\) c) \(\frac12\); the secant slopes approach the tangent slope at the origin.

- Direct substitution does not finish this limit; look for an equivalent expression valid near \(x=4\). - Think about how an expression containing a difference of square roots can be transformed so a factor cancels. - After simplifying, try substitution again.

1. Direct substitution gives \(\frac{0}{0}\), so rewrite the quotient in an equivalent form for \(x\ne 4\). 2. Multiply numerator and denominator by \(\sqrt{x}+2\): \(\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}=\frac{x-4}{(x-4)(\sqrt{x}+2)}=\frac{1}{\sqrt{x}+2}\). 3. Now let \(x\to4\): \(\frac{1}{\sqrt{4}+2}=\frac{1}{4}\).

\(\frac{1}{4}\)

- Use the slope formula between the origin and a general point \(P\) on the sine curve. - Read the dashed tangent line using two convenient grid points. - Compare the directions of the successive secants with the tangent as their endpoints move toward the origin.

1. The secant slope is \(\frac{\sin(x)-\sin(0)}{x-0}=\frac{\sin(x)}{x}\). 2. The tangent line passes through \((0,0)\) and \((1,1)\), so its slope is \(1\). 3. The displayed secant lines become closer to the tangent line as their second point moves toward the origin. Therefore, their slopes approach the tangent slope \(1\), so \(\lim_{x\to0}\frac{\sin(x)}{x}=1\).

1) \(\frac{\sin(x)}{x}\) 2) \(1\) 3) The secants approach the tangent, so \(\lim_{x\to0}\frac{\sin(x)}{x}=1\).