proof the sequential criterion for limits

The **Sequential Criterion for Limits** states that:

A function \( f \) has a limit \( L \) as \( x \) approaches \( c \) (written as \( \lim_{x \to c} f(x) = L \)) if and only if for every sequence \( \{ x_n \} \) that converges to \( c \) (with \( x_n \neq c \) for all \( n \)), the sequence \( \{ f(x_n) \} \) converges to \( L \).

In symbols:
\[
\lim_{x \to c} f(x) = L \quad \text{if and only if} \quad \lim_{n \to \infty} f(x_n) = L \text{ whenever } \lim_{n \to \infty} x_n = c \text{ and } x_n \neq c.
\]

### **Proof:**

#### (⇒) **If \( \lim_{x \to c} f(x) = L \), then for every sequence \( \{ x_n \} \) converging to \( c \), \( \{ f(x_n) \} \) converges to \( L \):**

Assume that \( \lim_{x \to c} f(x) = L \). This means that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that:
\[
0 < |x - c| < \delta \quad \text{implies} \quad |f(x) - L| < \epsilon.
\]

Let \( \{ x_n \} \) be any sequence such that \( \lim_{n \to \infty} x_n = c \) and \( x_n \neq c \) for all \( n \).

Since \( \lim_{n \to \infty} x_n = c \), for every \( \delta > 0 \), there exists \( N \) such that for all \( n \geq N \),
\[
0 < |x_n - c| < \delta.
\]

By the definition of the limit of \( f(x) \), we know that when \( 0 < |x - c| < \delta \), we have \( |f(x) - L| < \epsilon \). Therefore, for all \( n \geq N \),
\[
|f(x_n) - L| < \epsilon.
\]
This shows that \( \lim_{n \to \infty} f(x_n) = L \).

#### (⇐) **If for every sequence \( \{ x_n \} \) converging to \( c \), \( \{ f(x_n) \} \) converges to \( L \), then \( \lim_{x \to c} f(x) = L \):**

Now assume that for every sequence \( \{ x_n \} \) converging to \( c \) (with \( x_n \neq c \)), we have \( \lim_{n \to \infty} f(x_n) = L \).

To prove \( \lim_{x \to c} f(x) = L \), we must show that for every \( \epsilon > 0 \), there exists \( \delta > 0 \) such that:
\[
0 < |x - c| < \delta \quad \text{implies} \quad |f(x) - L| < \epsilon.
\]

Assume for contradiction that \( \lim_{x \to c} f(x) \neq L \). Then, there exists some \( \epsilon_0 > 0 \) such that for every \( \delta > 0 \), there exists some \( x \) with \( 0 < |x - c| < \delta \) and \( |f(x) - L| \geq \epsilon_0 \).

We can construct a sequence \( \{ x_n \} \) converging to \( c \) such that \( 0 < |x_n - c| < \frac{1}{n} \) and \( |f(x_n) - L| \geq \epsilon_0 \) for all \( n \). 

However, by the assumption of the sequential criterion, \( \lim_{n \to \infty} f(x_n) = L \), which contradicts the fact that \( |f(x_n) - L| \geq \epsilon_0 \) for all \( n \).

Thus, our assumption was wrong, and it must be true that \( \lim_{x \to c} f(x) = L \).

### **Conclusion:**
The sequential criterion is now proven: \( \lim_{x \to c} f(x) = L \) if and only if for every sequence \( \{ x_n \} \) converging to \( c \), \( \lim_{n \to \infty} f(x_n) = L \).

---

Would you like any more details, or do you have any questions?

Here are five related questions:
1. How is the sequential criterion applied to continuous functions?
2. Can the sequential criterion be used to prove the limit does not exist?
3. How does the sequential criterion relate to limits at infinity?
4. What is the relationship between the sequential criterion and the epsilon-delta definition of limits?
5. How can the sequential criterion be used to analyze discontinuities in functions?

**Tip:** When proving limits using the sequential criterion, always ensure that all sequences converging to the point in question are considered, not just a few specific ones.

Explore a detailed proof of the sequential criterion for limits. This criterion states that a function has a limit at a point if, for every sequence converging to that point, the function's values converge to the same limit. This concept is essential for understanding continuity and limits in advanced calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation