Algebra of Limits
Calculating limits from the definition is hard. Limit theorems allow us to build complex limits from simple ones. If \( \lim s_n = s \) and \( \lim t_n = t \), then limits behave "nicely" with respect to arithmetic operations.
Visualizing Linearity: \( \lim(s_n + t_n) = s + t \)
The red sequence \( s_n + t_n \) automatically converges to \( s + t \).
The Squeeze Theorem
Also known as the Sandwich Theorem. If a sequence is "squeezed" between two other sequences that converge to the same limit, it must also converge to that limit.
Theorem: Suppose \( a_n \le b_n \le c_n \) for all \( n \gt N \). If \( \lim a_n = \lim c_n = L \), then \( \lim b_n = L \).
Practice Problems
True/False
1. \( \lim (s_n t_n) = (\lim s_n)(\lim t_n) \) is true for all sequences.
False. This rule only applies if both \( \lim s_n \) and \( \lim t_n \) exist (converge).
2. If \( \lim s_n = \infty \) and \( \lim t_n = 0 \), then \( \lim (s_n t_n) = 0 \).
False. This is an indeterminate form \( \infty \cdot 0 \). The limit could be anything (0, \( \infty \), or a constant).
3. The Squeeze Theorem can be used to find the limit of \( \frac{\cos n}{n} \).
True. Since \( -1 \le \cos n \le 1 \), we have \( -\frac{1}{n} \le \frac{\cos n}{n} \le \frac{1}{n} \). Both bounds go to 0.
Fill in the Blank
1. If \( \lim s_n = s \) and \( \lim t_n = t \) with \( t \neq 0 \), then \( \lim (s_n/t_n) = \) ___.
Answer: \( s/t \)
2. To evaluate \( \lim \frac{3n^2+n}{5n^2-2} \), we should divide the numerator and denominator by ___.
Answer: \( n^2 \) (the highest power in the denominator)
3. If \( a_n \le s_n \le b_n \) and \( \lim a_n = \lim b_n = L \), then \( \lim s_n = \) ___.
Answer: \( L \)
Full Problems
1. Calculate \( \lim \frac{3n^2 - 5n}{n^2 + 2n + 1} \).
Divide numerator and denominator by \( n^2 \):
\( \lim \frac{3 - 5/n}{1 + 2/n + 1/n^2} = \frac{3 - 0}{1 + 0 + 0} = 3 \).
2. Calculate \( \lim (\sqrt{n^2+n} - n) \).
Multiply by the conjugate \( \frac{\sqrt{n^2+n} + n}{\sqrt{n^2+n} + n} \):
\( \frac{n^2+n - n^2}{\sqrt{n^2+n} + n} = \frac{n}{\sqrt{n^2+n} + n} \).
Divide by \( n \): \( \frac{1}{\sqrt{1+1/n} + 1} \to \frac{1}{\sqrt{1} + 1} = \frac{1}{2} \).
3. Calculate \( \lim \frac{\sin(n^2)}{n} \).
Use Squeeze Theorem:
\( -1 \le \sin(n^2) \le 1 \implies -\frac{1}{n} \le \frac{\sin(n^2)}{n} \le \frac{1}{n} \).
Since \( \lim (-1/n) = 0 \) and \( \lim (1/n) = 0 \), the limit is 0.
4. Calculate \( \lim \frac{2^n + 3^n}{3^n + 1} \).
Divide numerator and denominator by \( 3^n \) (the dominant term):
\( \lim \frac{(2/3)^n + 1}{1 + (1/3)^n} \).
Since \( |2/3| \lt 1 \) and \( |1/3| \lt 1 \), their powers go to 0.
Limit = \( \frac{0 + 1}{1 + 0} = 1 \).