Math Problem Statement

proof the sequential criterion for limits

Solution

The Sequential Criterion for Limits states that:

A function ff has a limit LL as xx approaches cc (written as limxcf(x)=L\lim_{x \to c} f(x) = L) if and only if for every sequence {xn}\{ x_n \} that converges to cc (with xncx_n \neq c for all nn), the sequence {f(xn)}\{ f(x_n) \} converges to LL.

In symbols: limxcf(x)=Lif and only iflimnf(xn)=L whenever limnxn=c and xnc.\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 limxcf(x)=L\lim_{x \to c} f(x) = L, then for every sequence {xn}\{ x_n \} converging to cc, {f(xn)}\{ f(x_n) \} converges to LL:

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

Let {xn}\{ x_n \} be any sequence such that limnxn=c\lim_{n \to \infty} x_n = c and xncx_n \neq c for all nn.

Since limnxn=c\lim_{n \to \infty} x_n = c, for every δ>0\delta > 0, there exists NN such that for all nNn \geq N, 0<xnc<δ.0 < |x_n - c| < \delta.

By the definition of the limit of f(x)f(x), we know that when 0<xc<δ0 < |x - c| < \delta, we have f(x)L<ϵ|f(x) - L| < \epsilon. Therefore, for all nNn \geq N, f(xn)L<ϵ.|f(x_n) - L| < \epsilon. This shows that limnf(xn)=L\lim_{n \to \infty} f(x_n) = L.

(⇐) If for every sequence {xn}\{ x_n \} converging to cc, {f(xn)}\{ f(x_n) \} converges to LL, then limxcf(x)=L\lim_{x \to c} f(x) = L:

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

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

Assume for contradiction that limxcf(x)L\lim_{x \to c} f(x) \neq L. Then, there exists some ϵ0>0\epsilon_0 > 0 such that for every δ>0\delta > 0, there exists some xx with 0<xc<δ0 < |x - c| < \delta and f(x)Lϵ0|f(x) - L| \geq \epsilon_0.

We can construct a sequence {xn}\{ x_n \} converging to cc such that 0<xnc<1n0 < |x_n - c| < \frac{1}{n} and f(xn)Lϵ0|f(x_n) - L| \geq \epsilon_0 for all nn.

However, by the assumption of the sequential criterion, limnf(xn)=L\lim_{n \to \infty} f(x_n) = L, which contradicts the fact that f(xn)Lϵ0|f(x_n) - L| \geq \epsilon_0 for all nn.

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

Conclusion:

The sequential criterion is now proven: limxcf(x)=L\lim_{x \to c} f(x) = L if and only if for every sequence {xn}\{ x_n \} converging to cc, limnf(xn)=L\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.

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Math Problem Analysis

Mathematical Concepts

Limits
Sequences
Convergence

Formulas

lim_{x → c} f(x) = L
lim_{n → ∞} x_n = c implies lim_{n → ∞} f(x_n) = L

Theorems

Sequential Criterion for Limits
Epsilon-Delta Definition of Limits

Suitable Grade Level

University Level (Advanced Calculus or Real Analysis)