Chapter 7: Problem 10
True, or false? As \(x \rightarrow \infty,\) \begin{equation} \begin{array}{ll}{\text { a. } \frac{1}{x+3}=o\left(\frac{1}{x}\right)} & {\text { b. } \frac{1}{x}+\frac{1}{x^{2}}=O\left(\frac{1}{x}\right)} \\\ {\text { c. } \frac{1}{x}-\frac{1}{x^{2}}=o\left(\frac{1}{x}\right)} & {\text { d. } 2+\cos x=O(2)} \\ {\text { e. } e^{x}+x=O\left(e^{x}\right)} & {\text { f. } x \ln x=o\left(x^{2}\right)} \\ {\text { g. } \ln (\ln x)=O(\ln x)} & {\text { h. } \ln (x)=o\left(\ln \left(x^{2}+1\right)\right)}\end{array} \end{equation}
Short Answer
Step by step solution
Understanding Little-o Notation
Understanding Big-O Notation
Analyzing Part (a)
Analyzing Part (b)
Analyzing Part (c)
Analyzing Part (d)
Analyzing Part (e)
Analyzing Part (f)
Analyzing Part (g)
Analyzing Part (h)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Little-o notation
- If \( f(x) = o(g(x)) \), it means that the function \( f(x) \) becomes negligibly small when compared to \( g(x) \) as \( x \to \infty \).
- Mathematically, this relationship is expressed through limits: \[ \lim_{{x \to \infty}} \frac{f(x)}{g(x)} = 0 \]
- So, no matter how large \( g(x) \) becomes, \( f(x) \) becomes even smaller in comparison, eventually approaching zero relative to \( g(x) \).
Big-O notation
- If \( f(x) = O(g(x)) \), it signifies that \( f(x) \) does not grow faster than \( g(x) \), or in simpler terms, it grows at the same rate or slower.
- This is formally defined as: there exist positive constants \( C \) and \( k \) such that \( |f(x)| \leq C|g(x)| \) for all \( x > k \).
Limits at infinity
- The key question is: What value does the function approach as \( x \rightarrow \infty \)?
- This helps us understand whether functions stabilize, grow indefinitely, or approach another pattern as \( x \) increases.