10. True, or false? As x→∞,
c. 1/x - 1/x² = o(1/x)

Verified step by step guidance

1. Recall the definition of the little-o notation: for functions f(x) and g(x), we say f(x) = o(g(x)) as x → ∞ if \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0 \).
2. Identify the functions in the problem: \( f(x) = \frac{1}{x} - \frac{1}{x^2} \) and \( g(x) = \frac{1}{x} \).
3. Form the ratio \( \frac{f(x)}{g(x)} = \frac{\frac{1}{x} - \frac{1}{x^2}}{\frac{1}{x}} \).
4. Simplify the ratio: \( \frac{\frac{1}{x} - \frac{1}{x^2}}{\frac{1}{x}} = \frac{1}{x} \cdot \frac{x}{1} - \frac{1}{x^2} \cdot \frac{x}{1} = 1 - \frac{1}{x} \).
5. Evaluate the limit as \( x \to \infty \): \( \lim_{x \to \infty} \left(1 - \frac{1}{x}\right) = 1 \). Since this limit is not zero, conclude that \( \frac{1}{x} - \frac{1}{x^2} \neq o\left(\frac{1}{x}\right) \).

Key Concepts

Little-o Notation

Little-o notation, written as f(x) = o(g(x)) as x→∞, means that f(x) becomes insignificant compared to g(x) in the limit. Formally, f(x)/g(x) → 0 as x→∞, indicating f(x) grows much slower than g(x). It is used to describe the relative growth rates of functions.

Asymptotic Behavior of Rational Functions

As x approaches infinity, terms like 1/x and 1/x² approach zero, but at different rates. Specifically, 1/x² tends to zero faster than 1/x, which affects how their differences behave asymptotically. Understanding these rates is key to comparing functions using little-o notation.

Limit of a Quotient for Comparing Growth Rates

To determine if f(x) = o(g(x)), evaluate the limit of f(x)/g(x) as x→∞. If the limit is zero, f(x) is little-o of g(x). This method helps verify statements about asymptotic dominance, such as whether 1/x - 1/x² = o(1/x).