Exercises
Mathematics I
Week 7 - Sequence and Limits

Week 7 - Sequence and Limits


Graded Assignment

1. Multiple Choice/Statement Analysis

Statements about sequences:

  1. Statement: If an{a_n} and bn{b_n} are two sequences of real numbers, then an+bn{a_n + b_n} is a convergent sequence.
    • Counterexample: Let an=1a_n = 1 for all nn, bn=−1b_n = -1 for all nn. Both converge, but an+bn=0a_n + b_n = 0 for all nn, which is convergent. However, the PDF says “option 1 is not correct,” which may refer to a different statement or a misinterpretation. The given explanation is not clear, but the PDF concludes: “Hence option 1 is not correct.”
  2. Statement: If an{a_n} is an increasing sequence, then (−1)nan{(-1)^n a_n} is a decreasing sequence.
    • Counterexample: an=na_n = n for all nn. Then (−1)nan{(-1)^n a_n} is not decreasing.
    • Solution: Hence option 2 is not correct.
  3. Statement: If an→a{a_n} \to a, bn→b{b_n} \to b, and both a,ba, b are non-zero, then anbn→ab{a_n b_n} \to ab must be non-zero.
    • Solution: This is correct.
    • Conclusion: Option 3 is correct.
  4. Statement: If an→a{a_n} \to a and bn→a{b_n} \to a, then an−bn→0{a_n - b_n} \to 0.
    • Solution: This is correct.
    • Conclusion: Option 4 is correct.
  5. Statement: If a sequence is divergent, then any subsequence is also divergent.
    • Counterexample: Let an=na_n = n if nn is odd, an=1a_n = 1 if nn is even. an{a_n} is divergent, but a2n{a_{2n}} is constant and hence convergent.
    • Conclusion: Option 5 is not correct.

2. Function Type Matching

Match the following functions to their types:

  • i) f(x)=3ln⁡x−2f(x) = 3 \ln x - 2
    • Type: Logarithmic function (d)
  • ii) f(x)=10−4xf(x) = 10 - 4x
    • Type: Linear function (c)
  • iii) f(x)=2x+7f(x) = 2^x + 7
    • Type: Exponential function (a)
  • iv) f(x)=x2−4x+4f(x) = x^2 - 4x + 4
    • Type: Quadratic function (b)

3. Limit and Function Behavior

Given a graph and options:

  • Option 2: lim⁡x→0+f(x)=0=lim⁡x→0−f(x)\lim_{x \to 0^+} f(x) = 0 = \lim_{x \to 0^-} f(x). Hence option 2 is correct.
  • Option 5: At x=πx = \pi and x=−πx = -\pi there are no sharp corners at the given curve. So, option 5 is correct.
  • Interval [−0.5π,0.5π][-0.5\pi, 0.5\pi]: The function is oscillatory (neither monotonically increasing nor monotonically decreasing).

4. Limit Calculations

  • a) lim⁡x→0log⁡(1+x)sin⁡x=1\lim_{x \to 0} \frac{\log(1+x)}{\sin x} = 1
  • b) lim⁡x→0sin⁡5xx=5\lim_{x \to 0} \frac{\sin 5x}{x} = 5
  • c) lim⁡x→0ex/2−1sin⁡2x=14\lim_{x \to 0} \frac{e^{x/2} - 1}{\sin 2x} = \frac{1}{4}

5. Function Comparison

Given f(x)>g(x)f(x) > g(x) for all x≥x0x \ge x_0 and f(x)≤g(x)f(x) \le g(x) for x≤x0x \le x_0:

  • Option 2: f(x)f(x) and g(x)g(x) will never intersect for x>x0x > x_0 is incorrect, since f(x0)=g(x0)f(x_0) = g(x_0). Solution: Option 2 is incorrect.

6. Sequence Limit

Given:

an=12n23n+5−4n2+7n+3 a_n = \frac{12n^2}{3n+5} - \frac{4n^2 + 7}{n+3}

Simplify and find the limit as n→∞n \to \infty:

an=16n2−21n−353n2+14n+15  ⟹  lim⁡n→∞an=163 a_n = \frac{16n^2 - 21n - 35}{3n^2 + 14n + 15} \implies \lim_{n \to \infty} a_n = \frac{16}{3}

7. Function Limit and Monotonicity

Given:

R(w)=50ew10+ew R(w) = \frac{50e^w}{10 + e^w}
  • Monotonicity: R(w)R(w) is increasing.
  • Limit as w→∞w \to \infty: R(w)→50R(w) \to 50
  • Minimum r∈Zr \in \mathbb{Z} such that R(w)<rR(w) < r for all ww: r=50r = 50

8. Advanced Limit Calculation

Given:

lim⁡n→∞en!n[log⁡(1+6n)−e1/n−1(2πn)n]=5 \lim_{n \to \infty} e \sqrt[n]{n!} \left[ \log(1 + \frac{6}{n}) - \frac{e^{1/n} - 1}{\sqrt[n]{(2\pi n)}} \right] = 5

9. Curve Analysis

Both curve 1 and curve 2 have sharp corners at the origin (0,0). Hence, at the origin these two curves do not have tangents at the origin.

10. Sequence Sum and Limit

Given:

an=9+15+21+⋯+3(2n−1)n2 a_n = \frac{9 + 15 + 21 + \dots + 3(2n-1)}{n^2}

Simplify and find the limit as n→∞n \to \infty:

an=3(n2−1)n2=3(1−1n2)  ⟹  lim⁡n→∞an=3 a_n = \frac{3(n^2 - 1)}{n^2} = 3\left(1 - \frac{1}{n^2}\right) \implies \lim_{n \to \infty} a_n = 3

11. Limit Involving Floor Function

Given:

5lim⁡x→3+[x]−3lim⁡x→1−[x] 5 \lim_{x \to 3^+} [x] - 3 \lim_{x \to 1^-} [x]

Solution:

=5×3−3×0=15 = 5 \times 3 - 3 \times 0 = 15

12–14. Algorithm Error Estimation

12. Error in estimation by Algorithm 1:

an=n2+5n6n2+1,lim⁡n→∞an=16≈0.166 a_n = \frac{n^2 + 5n}{6n^2 + 1}, \quad \lim_{n \to \infty} a_n = \frac{1}{6} \approx 0.166

Error in estimation by Algorithm 2:

bn=18+(−1)nn,lim⁡n→∞bn=18=0.125 b_n = \frac{1}{8} + \frac{(-1)^n}{n}, \quad \lim_{n \to \infty} b_n = \frac{1}{8} = 0.125

Error in estimation by Algorithm 3:

cn=en+47en,lim⁡n→∞cn=17≈0.143 c_n = \frac{e^n + 4}{7e^n}, \quad \lim_{n \to \infty} c_n = \frac{1}{7} \approx 0.143

Conclusion:

  • Maximum error: Algorithm 1
  • Minimum error: Algorithm 2

13. Error in estimation by new algorithm:

lim⁡(an−bn)=16−18=124 \lim (a_n - b_n) = \frac{1}{6} - \frac{1}{8} = \frac{1}{24}

The error in estimation using the new algorithm is less than the error in estimation using any of the Algorithm 1, Algorithm 2 and Algorithm 3.

14. Limit calculation:

cn′=ne18n−n c_n' = n e^{\frac{1}{8n}} - n lim⁡n→∞cn′=18=0.125 \lim_{n \to \infty} c_n' = \frac{1}{8} = 0.125

This covers all questions and solutions from the PDF1.

⁂

  1. Week-7-Sequence-and-Limits.pdf ↩︎

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