# 3.2 Asymptotics

## Practice with Order of Growth

For each pair of $$f(n)$$ and $$g(n)$$ below, determine which of the listed relations apply.

1. $$f(n) = log_{3} n$$ and $$g(n) = log_{7} n$$

Exercise 1

$$\lim_{n \rightarrow \infty} |\frac{f(n)}{g(n)}| = \lim_{n \rightarrow \infty} \frac{ln(n) / ln(3)}{ln(n) / ln(7)} = \frac{ln(7)}{ln(3)}$$
2. $$f(n) = 0$$ and $$g(n) = 33$$

Exercise 2

$$\lim_{n \rightarrow \infty} |\frac{f(n)}{g(n)}| = \lim_{n \rightarrow \infty} \frac{0}{33} = 0$$
3. $$f(n) = 1 + cos(\frac{\pi n}{2})$$ and $$g(n) = 1 + sin(\frac{\pi n}{2})$$

Exercise 3

$$\frac{f(n)}{g(n)} = 0$$ for $$n \equiv 1 (\text{mod } 4)$$, and $$\frac{g(n)}{f(n)} = 0$$ for $$n \equiv 0 (\text{mod } 4)$$, so the 2 functions and their quotient never converge to some limit.
4. $$f(n) = 1.01^{n}$$ and $$g(n) = n^{100}$$

Exercise 4

$$\lim_{n \rightarrow \infty} |\frac{f(n)}{g(n)}| = \lim_{n \rightarrow \infty} \frac{1.01^{n}}{n^{100}} = \lim_{n \rightarrow \infty} \frac{1.01^{n} ln(1.01)}{100 \cdot n^{99}} = ... = \lim_{n \rightarrow \infty} \frac{1.01^{n} ln(1.01)^{100}}{100!} = \infty$$