Chapter 2: Problem 95
Use formal definitions to prove the limit statements. $$\lim _{x \rightarrow 0} \frac{-1}{x^{2}}=-\infty$$
Short Answer
Expert verified
The proof shows \( \frac{-1}{x^2} \to -\infty \) as \( x \to 0 \).
Step by step solution
01
Understand the Formal Definition of Limits
To prove that \( \lim_{x \rightarrow 0} \frac{-1}{x^2} = -\infty \), we need to demonstrate that for every large negative number \( M \), there exists a \( \delta > 0 \) such that if \( 0 < |x| < \delta \), then \( \frac{-1}{x^2} < M \). This interpretation indicates the function goes to \(-\infty\) as \( x \) approaches \( 0 \).
02
Determine the Function Behavior
Consider the function \( f(x) = \frac{-1}{x^2} \). As \( x \) approaches 0 (but \( x eq 0 \)), \( \frac{1}{x^2} \) becomes very large. Since \( -1 \) is multiplied with this large positive value, the function becomes a very large negative value.
03
Choose an Appropriate \( \delta \)
For any given negative number \( M \), consider \( \delta = \frac{1}{\sqrt{-M}} \). This choice ensures that when \( 0 < |x| < \delta \), \( \frac{1}{x^2} > -M \), meaning \( \frac{-1}{x^2} < M \).
04
Validate the Chosen \( \delta \)
Check if the chosen \( \delta = \frac{1}{\sqrt{-M}} \) satisfies the condition. Since \( x^2 < \delta^2 = \frac{1}{-M} \), it implies \( \frac{1}{x^2} > -M \), thus \( \frac{-1}{x^2} < M \) which verifies the initial statement.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Formal Definition of Limits
Understanding the formal definition of limits is a foundational concept in calculus. It involves determining how a function behaves as its input approaches a certain value. When we say that the limit of a function \(f(x)\) as \(x\) approaches a point \((c)\) is equal to \(L\) (written as \(\lim_{x \rightarrow c} f(x) = L\)), it means:
- For every number \(\epsilon > 0\), there's a number \(\delta > 0\) such that whenever \(|x - c| < \delta\), we have \(|f(x) - L| < \epsilon\).
Infinite Limits
Infinite limits are a special type of limit that describe the behavior of a function as it grows without bound near a point. When we say a function approaches infinity or negative infinity, it means:
- For every large positive number \(M\) (or for negative infinity, a large negative number), we find there's a \(\delta > 0\) so that \(0 < |x-c| < \delta\) ensures \(f(x) > M\) (or \(f(x) < M\) for negative infinity).
Delta-Epsilon Definition
The delta-epsilon (\(\delta-\epsilon\)) definition is an essential tool for proving limits, ensuring the continuity of the calculus framework. Here is how it works:
- The \(\epsilon\) represents how close you want the output of the function \(f(x)\) to get to the limit \(L\).
- The \(\delta\), in turn, dictates how close \(x\) must be to \(c\) to meet the \(\epsilon\)-closeness requirement of \(f(x)\).
- For a function \(f(x)\) and limit \[L\] as \[x ightarrow c\], choose \[\epsilon > 0\], then find \[\delta > 0\] such that \[0 < |x-c| < \delta\Rightarrow |f(x) - L| < \epsilon\].