Problem 35 - Calculus of a Single Variable

Question

Finding a \(\delta\) for a Given \(\varepsilon\) In Exercises \(33-36\) , find the limit \(L\) . Then find \(\delta>0\) such that \(|f(x)-L|<0.01\) whenever \(0<|x-c|<\delta .\) $$ \lim _{x \rightarrow 2}\left(x^{2}-3\right) $$

Step-by-Step Solution

Answer

The limit is 1. Any delta less than 0.0025 will satisfy the given condition.

Step 1: Find the limit

The limit of the function \(x^2 - 3\) as \(x\) approaches 2 can be found by replacing \(x\) with 2. So, \(L = (2^2) - 3 = 1.\)

Step 2: Find Delta

The inequality \(|f(x) - L| < 0.01\) needs to be solved to find delta. Substitute \(f(x)\) and \(L\) with their values. This results in \(|x^2 - 3 - 1| < 0.01\) or \(|(x - 2)(x + 2)| < 0.01\). The fact that \(0 < |x - 2| < \delta\) helps us to form \(|x + 2| \approx 4\) for \(\delta\) sufficiently small. Hence, \(|(x - 2)(4)| < 0.01\) leading to \(4\delta < 0.01\). Solving for \(\delta\) gives us \(\delta < 0.0025\). Thus any \(\delta\) less than \(0.0025\) will satisfy the given condition.

Key Concepts

Limit

A limit is an essential concept in calculus. It describes the value that a function approaches as the input "gets closer" to some value. For instance, when a problem asks for \( \lim_{x \rightarrow 2} (x^2 - 3) \), it wants to know what \( x^2 - 3 \) becomes as \( x \) approaches 2.
In the example given, by plugging \( x = 2 \) into the expression \( x^2 - 3 \), we calculate \( 2^2 - 3 = 1 \).
This result, 1, is the limit. The process involves substituting \( x \) with a number very close to, but not exactly equal to, 2, and observing the function value's behavior.

Limits can help us understand function behavior at points where it's not clearly defined or straightforward. It forms the foundation for concepts like derivatives and integrals.

Continuity

Continuity in mathematics describes a situation where a function doesn't have any interruptions, jumps, or "breaks" at a certain point. A function is continuous at a point if its limit equals its actual function value there.
In terms of the epsilon-delta definition, a function \( f(x) \) is continuous at \( x = c \) if for any \( \varepsilon > 0 \), there's a \( \delta > 0 \) so that whenever \( 0 < |x-c| < \delta \), then \( |f(x) - L| < \varepsilon \).
In the provided exercise, we're ensuring that the function \( x^2 - 3 \) remains uninterrupted around \( x = 2 \) by finding a suitable \( \delta \). While it may initially seem a bit abstract, continuity ensures smoothness in function behavior, making it easier to predict how the function behaves around a given point.
Continuity is a fundamental quality that simplifies working with functions and is crucial for defining derivatives.

Inequality

An inequality denotes a relationship between two expressions. It suggests that one expression is less than, greater than, or not equal to another. In the epsilon-delta context, it appears as: \( |f(x) - L| < \varepsilon \).
This representation involves the absolute value, implying that the exact direction (above or below) doesn't matter as long as the difference falls within a specified bound, in this case, smaller than \( 0.01 \).
The exercise used the inequality \( |(x - 2)(x + 2)| < 0.01 \) to determine \( \delta \), restricting how far \( x \) can strays from 2 while still meeting condition \( |f(x) - L| < 0.01 \).
Solving inequalities often requires manipulation to isolate the variable of interest, as demonstrated when solving for \( \delta < 0.0025 \). Understanding inequalities helps in determining the precision and accuracy needed around points, which is vital in calculus and various practical applications.