The Precise Definition of a Limit: Fresh Take

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The Precise Definition of a Limit: Fresh Take

Use the epsilon-delta method to determine the limit of a function
Explain the epsilon-delta definitions of one-sided limits and infinite limits

Epsilon-Delta Definition of the Limit

The Main Idea

Formal Definition:
- For f(x) defined near a, limx→af(x)=L if:
  - For every ε>0, there exists δ>0 such that:
    - If $0 < |x - a| < \delta$ , then $|f(x) - L| < \varepsilon$
Geometric Interpretation:
- $\varepsilon$ : Vertical distance from $L$
- $\delta$ : Horizontal distance from $a$
- As $\varepsilon$ gets smaller, $\delta$ typically needs to be smaller
Key Equivalences:
- $|f(x) - L| < \varepsilon$ is equivalent to $L - \varepsilon < f(x) < L + \varepsilon$
- $0 < |x - a| < \delta$ is equivalent to $a - \delta < x < a + \delta$ and $x \neq a$

Complete the proof that $\underset{x\to 2}{\lim}(3x-2)=4$ by filling in the blanks.

Let _____.

Choose $\delta =$ ________.

Assume $0<|x-\text{___}|<\delta$ .

Thus, $|\text{________}-\text{___}| =|\text{_________}| = |\text{___}||\text{_________}| = \text{___} \, |\text{_______}| < \text{______} = \text{_______} = \varepsilon$ .

Therefore, $\underset{x\to 2}{\lim}(3x-2)=4$ .

Let $\varepsilon >0$ ; choose $\delta =\frac{\varepsilon}{3}$ ; assume $0<|x-2|<\delta$ .

Thus, $|(3x-2)-4|=|3x-6|=|3| \cdot |x-2|<3 \cdot \delta =3 \cdot (\varepsilon/3)=\varepsilon$ .

Therefore, $\underset{x\to 2}{\lim}3x-2=4$ .

Find $\delta$ corresponding to $\varepsilon >0$ for a proof that $\underset{x\to 9}{\lim}\sqrt{x}=3$ .

Choose $\delta =\text{min}\{9-(3-\varepsilon)^2,(3+\varepsilon)^2-9\}$ .

Complete the proof that $\underset{x\to 1}{\lim}x^2=1$ .

Let $\varepsilon >0$ ; choose $\delta =\text{min}\{1,\frac{\varepsilon}{3}\}$ ; assume $0<|x-1|<\delta$ .

Since $|x-1|<1$ , we may conclude that [latex]-1

$|x^2-1|=|x-1| \cdot |x+1|<\varepsilon/3 \cdot 3=\varepsilon$

Advanced Applications of the Epsilon-Delta Definition: Proofs, Non-Existence, and Algebraic Calculations

The Main Idea

Triangle Inequality:
- For any real numbers $a$ and $b$ , $|a+b| \leq |a| + |b|$
- 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 $\varepsilon > 0$ that works for all $\delta > 0$
Algebraic Approach to Finding Deltas:
- Solve $|f(x) - L| < \varepsilon$ for $x$
- Find $\delta$ that ensures $|x - x_0| < \delta$ implies $|f(x) - L| < \varepsilon$

Find an open interval about $x_0$ on which the inequality $|f(x)-L| < 0$ holds. Then give the largest value $\delta > 0$ such that for all $x$ satisfying $0 < |x-x_0| < \delta$ the inequality $|f(x)-L| < \varepsilon$ holds.

$f(x)=\sqrt{x+4}, \,\, L=3, \,\, x_0=5, \,\, \varepsilon=1$

First we will need to start with the inequality

$|f(x)-L| < \varepsilon$ and plug in our numbers. Then we will solve for

$x$ .

$|\sqrt(x+4) - 3| < \varepsilon$

$|\sqrt{x+4} - 3| < 1$

$-1 < \sqrt{x+4} - 3 < 1$

$2 < \sqrt{x+4} < 4$

$4 < x + 4 < 16$

$0 < x + 4 < 12$

Therefore, the interval is $(0,12)$ . For the second answer, we will start with $0 < |x-x_0| < \delta$ . We will plug in our value and solve:

$|x-5| < \delta$

$-\delta < x-5 < \delta$

$5-\delta < x < 5+\delta$

Now we will set each piece equal to the endpoints we found above.

$5-\delta=0$ and

$5+\delta=12$

After solving we will get $\delta=5 \text{ and }\delta=7$ . Since the question is asking for the smallest interval, we choose the smaller number. Therefore the answer is $\delta=5$ .

One-Sided and Infinite Limits

The Main Idea

One-Sided Limits:
- Limit from the right: $\lim_{x \to a^+} f(x) = L$
- Limit from the left: $\lim_{x \to a^-} f(x) = L$
- Modifications to standard epsilon-delta definition
Infinite Limits:
- Positive infinite limit: $\lim_{x \to a} f(x) = +\infty$
- Negative infinite limit: $\lim_{x \to a} f(x) = -\infty$
- Replace $\varepsilon$ with $M$ for arbitrarily large values

Formal Definitions

Limit from the Right: For every $\varepsilon > 0$ , there exists $\delta > 0$ such that: If $0 < x - a < \delta$ , then $|f(x) - L| < \varepsilon$
Limit from the Left: For every $\varepsilon > 0$ , there exists $\delta > 0$ such that: If $0 < a - x < \delta$ , then $|f(x) - L| < \varepsilon$
Positive Infinite Limit: For every $M > 0$ , there exists $\delta > 0$ such that: If $0 < |x - a| < \delta$ , then $f(x) > M$
Negative Infinite Limit: For every $M > 0$ , there exists $\delta > 0$ such that: If $0 < |x - a| < \delta$ , then $f(x) < -M$

Find $\delta$ corresponding to $\varepsilon$ for a proof that $\underset{x\to 1^-}{\lim}\sqrt{1-x}=0$ .

$\delta =\varepsilon^2$

Prove that $\lim_{x \to 4^+} \sqrt{x-4} = 0$ .

Let $\varepsilon > 0$ . Choose $\delta = \varepsilon^2$ . Assume $0 < x - 4 < \delta$ .

Then:

$\begin{array}{rcl} 0 < x - 4 < \delta &=& \varepsilon^2 \ 0 < \sqrt{x-4} < \varepsilon \ |\sqrt{x-4} - 0| < \varepsilon \end{array}$

Therefore, $\lim_{x \to 4^+} \sqrt{x-4} = 0$ .