Problem 13
Question
Prove the given limit using an \(\varepsilon-\delta\) proof. $$ \lim _{x \rightarrow 1} \frac{1}{x}=1 $$
Step-by-Step Solution
Verified Answer
Since \(|x - 1| < \delta\) leads to \(\left| \frac{1}{x} - 1 \right| < \varepsilon\), the limit is proven.
1Step 1: Understanding the Limit Statement
We need to prove using an \(\varepsilon-\delta\) proof that \(\lim _{x \rightarrow 1} \frac{1}{x} = 1\). This means that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(|x - 1| < \delta\) then \(\left| \frac{1}{x} - 1 \right| < \varepsilon\).
2Step 2: Express the Absolute Difference
Consider the absolute difference \(\left| \frac{1}{x} - 1 \right| = \left| \frac{1-x}{x} \right| = \frac{|1-x|}{|x|}\). To ensure this is less than \(\varepsilon\), choose \(\delta\) appropriately.
3Step 3: Propose a Bound for \(x\)
Assume \(|x - 1| < \delta\), which implies \(1-\delta < x < 1+\delta\). To have \(x\) not getting too close to zero (since that would make \(\frac{1}{x}\) undefined or large), let's propose that \(\delta < 0.5\). This gives \(0.5 < x < 1.5\).
4Step 4: Bound \(|x|\) Away from Zero
From the inequality \(0.5 < x < 1.5\), we have \(|x| \geq 0.5\) since the smallest positive \(x\) can be is 0.5. This bound helps control the troubling behavior near \(x = 0\).
5Step 5: Inequality Manipulation
We now have \(\frac{|1 - x|}{|x|} < \varepsilon\). Given \(|x| \geq 0.5\), it follows that:\[|1 - x| < \varepsilon |x|\leq \varepsilon \cdot 1.5.\]Chose \(\delta\) so that \(\delta \leq \varepsilon \cdot 0.5\).
6Step 6: Verify the Conditions with \(\delta\)
If \(\delta = \min\{0.5, \varepsilon \cdot 0.5\}\) is chosen, then for \(|x - 1| < \delta\), it ensures \(|x| \geq 0.5\) and:\[\frac{|1-x|}{|x|} < \varepsilon.\]This satisfies the \(\varepsilon-\delta\) condition.
Key Concepts
Understanding Limits in CalculusGrasping Continuity Through LimitsReal Analysis and Rigorous Proofs
Understanding Limits in Calculus
Limits are foundational to calculus. They describe how a function behaves as the input approaches a certain value. The notation \( \lim_{x \to a} f(x) = L \) implies that as \( x \) approaches \( a \), the function \( f(x) \) gets arbitrarily close to \( L \). This is crucial because it allows us to understand function behaviors even at values where they might not be explicitly defined.
In our exercise, the limit we are considering is \( \lim_{x \to 1} \frac{1}{x} = 1 \). This signifies that as \( x \) gets closer to 1 from both sides, the value of \( \frac{1}{x} \) approaches 1.
In our exercise, the limit we are considering is \( \lim_{x \to 1} \frac{1}{x} = 1 \). This signifies that as \( x \) gets closer to 1 from both sides, the value of \( \frac{1}{x} \) approaches 1.
- The role of limits involves exploring the "approaching behavior" rather than actual substitution.
- It helps in finding derivatives and evaluating integrals, which are the core operations in calculus.
Grasping Continuity Through Limits
Continuity means that a function doesn’t have any jumps, breaks, or infinite holes. A function is continuous at a point \( a \) if \( \lim_{x \to a} f(x) = f(a) \).
This connects seamlessly with the concept of limits, as they help us define continuity at every point within a function.
In our example, showing \( \lim_{x \to 1} \frac{1}{x} = 1 \) using an \( \varepsilon-\delta \) proof also speaks about continuity around \( x=1 \). What it does is:
This connects seamlessly with the concept of limits, as they help us define continuity at every point within a function.
In our example, showing \( \lim_{x \to 1} \frac{1}{x} = 1 \) using an \( \varepsilon-\delta \) proof also speaks about continuity around \( x=1 \). What it does is:
- Verifies the mathematical definition of a limit, implying the function is continuous around that point.
- Ensures the function value and the predicted limit match as \( x \) approaches 1.
Real Analysis and Rigorous Proofs
Real Analysis might seem daunting, but it's simply the in-depth study of real-valued sequences and their limits. It lays out the rules and justifications for why calculus works.
In the case of our \( \varepsilon-\delta \) proof for \( \lim_{x \to 1} \frac{1}{x} = 1 \), we're applying real analysis to establish certainty. Real analysis doesn’t take shortcuts; every statement must be proven.
In the case of our \( \varepsilon-\delta \) proof for \( \lim_{x \to 1} \frac{1}{x} = 1 \), we're applying real analysis to establish certainty. Real analysis doesn’t take shortcuts; every statement must be proven.
- It develops the idea of limits and continuity into a solid framework.
- Breaks down complex calculus concepts into logical, provable steps.
- Ensures mathematical results are foolproof by demanding proofs from basic axioms or previously established results.
Other exercises in this chapter
Problem 13
Evaluate the given limits of the piecewise defined functions \(f\). \(f(x)=\left\\{\begin{array}{cl}x+1 & x \leq 1 \\ x^{2}-5 & x>1\end{array}\right.\) (a) \(\l
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Use the following information to evaluate the given limit, when possible. If it is not possible to determine the limit, state why not. $$ \begin{array}{ll} *\li
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Approximate the given limits both numerically and graphically. $$ \begin{array}{l} \lim _{x \rightarrow 2} f(x), \text { where } \\ f(x)=\left\\{\begin{array}{c
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Evaluate the given limits using the graph of the function. \(f(x)=2^{x}+10\) (a) \(\lim _{x \rightarrow-\infty} f(x)\) (b) \(\lim _{x \rightarrow \infty} f(x)\)
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