- Use the epsilon-delta method to determine the limit of a function
- Explain the epsilon-delta definitions of one-sided limits and infinite limits
- Demonstrate how limit rules work using the epsilon-delta approach
Epsilon-Delta Definition of the Limit
The Main Idea
- Formal Definition:
- For [latex]f(x)[/latex] defined near [latex]a[/latex], [latex]\lim_{x \to a} f(x) = L[/latex] if:
- For every [latex]\varepsilon > 0[/latex], there exists [latex]\delta > 0[/latex] such that:
- If [latex]0 < |x - a| < \delta[/latex], then [latex]|f(x) - L| < \varepsilon[/latex]
- For every [latex]\varepsilon > 0[/latex], there exists [latex]\delta > 0[/latex] such that:
- For [latex]f(x)[/latex] defined near [latex]a[/latex], [latex]\lim_{x \to a} f(x) = L[/latex] if:
- Geometric Interpretation:
- [latex]\varepsilon[/latex]: Vertical distance from [latex]L[/latex]
- [latex]\delta[/latex]: Horizontal distance from [latex]a[/latex]
- As [latex]\varepsilon[/latex] gets smaller, [latex]\delta[/latex] typically needs to be smaller
- Key Equivalences:
- [latex]|f(x) - L| < \varepsilon[/latex] is equivalent to [latex]L - \varepsilon < f(x) < L + \varepsilon[/latex]
- [latex]0 < |x - a| < \delta[/latex] is equivalent to [latex]a - \delta < x < a + \delta[/latex] and [latex]x \neq a[/latex]
Complete the proof that [latex]\underset{x\to 2}{\lim}(3x-2)=4[/latex] by filling in the blanks.
Let _____.
Choose [latex]\delta =[/latex] ________.
Assume [latex]0<|x-\text{___}|<\delta[/latex].
Thus, [latex]|\text{________}-\text{___}| =|\text{_________}| = |\text{___}||\text{_________}| = \text{___} \, |\text{_______}| < \text{______} = \text{_______} = \varepsilon[/latex].
Therefore, [latex]\underset{x\to 2}{\lim}(3x-2)=4[/latex].
Find [latex]\delta[/latex] corresponding to [latex]\varepsilon >0[/latex] for a proof that [latex]\underset{x\to 9}{\lim}\sqrt{x}=3[/latex].
Complete the proof that [latex]\underset{x\to 1}{\lim}x^2=1[/latex].
Let [latex]\varepsilon >0[/latex]; choose [latex]\delta =\text{min}\{1,\frac{\varepsilon}{3}\}[/latex]; assume [latex]0<|x-1|<\delta[/latex].
Since [latex]|x-1|<1[/latex], we may conclude that [latex]-1
Advanced Applications of the Epsilon-Delta Definition: Proofs, Non-Existence, and Algebraic Calculations
The Main Idea
- Triangle Inequality:
- For any real numbers [latex]a[/latex] and [latex]b[/latex], [latex]|a+b| \leq |a| + |b|[/latex]
- Key tool in epsilon-delta proofs
- Proving Limit Laws:
- Use epsilon-delta definition to prove properties of limits
- Example: Sum of limits is limit of sum
- Non-Existence of Limits:
- A limit doesn’t exist if no real number satisfies the epsilon-delta definition
- Requires finding an [latex]\varepsilon > 0[/latex] that works for all [latex]\delta > 0[/latex]
- Algebraic Approach to Finding Deltas:
- Solve [latex]|f(x) - L| < \varepsilon[/latex] for [latex]x[/latex]
- Find [latex]\delta[/latex] that ensures [latex]|x - x_0| < \delta[/latex] implies [latex]|f(x) - L| < \varepsilon[/latex]
Find an open interval about [latex]x_0[/latex] on which the inequality [latex]|f(x)-L| < 0[/latex] holds. Then give the largest value [latex]\delta > 0[/latex] such that for all [latex]x[/latex] satisfying [latex]0 < |x-x_0| < \delta[/latex] the inequality [latex]|f(x)-L| < \varepsilon[/latex] holds.
One-Sided and Infinite Limits
The Main Idea
- One-Sided Limits:
- Limit from the right: [latex]\lim_{x \to a^+} f(x) = L[/latex]
- Limit from the left: [latex]\lim_{x \to a^-} f(x) = L[/latex]
- Modifications to standard epsilon-delta definition
- Infinite Limits:
- Positive infinite limit: [latex]\lim_{x \to a} f(x) = +\infty[/latex]
- Negative infinite limit: [latex]\lim_{x \to a} f(x) = -\infty[/latex]
- Replace [latex]\varepsilon[/latex] with [latex]M[/latex] for arbitrarily large values
Formal Definitions
- Limit from the Right: For every [latex]\varepsilon > 0[/latex], there exists [latex]\delta > 0[/latex] such that: If [latex]0 < x - a < \delta[/latex], then [latex]|f(x) - L| < \varepsilon[/latex]
- Limit from the Left: For every [latex]\varepsilon > 0[/latex], there exists [latex]\delta > 0[/latex] such that: If [latex]0 < a - x < \delta[/latex], then [latex]|f(x) - L| < \varepsilon[/latex]
- Positive Infinite Limit: For every [latex]M > 0[/latex], there exists [latex]\delta > 0[/latex] such that: If [latex]0 < |x - a| < \delta[/latex], then [latex]f(x) > M[/latex]
- Negative Infinite Limit: For every [latex]M > 0[/latex], there exists [latex]\delta > 0[/latex] such that: If [latex]0 < |x - a| < \delta[/latex], then [latex]f(x) < -M[/latex]
Find [latex]\delta[/latex] corresponding to [latex]\varepsilon[/latex] for a proof that [latex]\underset{x\to 1^-}{\lim}\sqrt{1-x}=0[/latex].
Prove that [latex]\lim_{x \to 4^+} \sqrt{x-4} = 0[/latex].