Answer
We have shown that if the limit of $f(x)$ as $x$ approaches $c$ exists and is equal to both $L_1$ and $L_2$, then $L_1$ must be equal to $L_2$. Therefore, if the limit exists, it must be unique.
Final Answer: The final answer is $\boxed{L_1 = L_2}$
Work Step by Step
Concept: Uniqueness of Limits
The concept of the uniqueness of a limit states that if the limit of a function $f(x)$ as $x$ approaches a certain value $c$ exists, then this limit must be a single, unique value. In simpler terms, a function cannot approach two different values simultaneously as its input approaches a specific point. This is a fundamental property of limits in calculus and real analysis.
Step-by-Step Solution
We will use the epsilon-delta definition of a limit to prove the uniqueness of the limit.
1. Assume Two Limits Exist:
Suppose that the limit of $f(x)$ as $x$ approaches $c$ exists and is equal to two different values, say $L_1$ and $L_2$. This means:
$\lim_{x \to c} f(x) = L_1$
$\lim_{x \to c} f(x) = L_2$
We want to show that this assumption leads to $L_1 = L_2$.
2. Apply the Epsilon-Delta Definition for Each Limit:
By the definition of a limit, for any $\epsilon > 0$, there exists a $\delta_1 > 0$ such that if $0 < |x - c| < \delta_1$, then $|f(x) - L_1| < \epsilon/2$.
Similarly, for the same $\epsilon > 0$, there exists a $\delta_2 > 0$ such that if $0 < |x - c| < \delta_2$, then $|f(x) - L_2| < \epsilon/2$.
We choose $\epsilon/2$ here for convenience in the subsequent steps.
3. Choose a Common Interval:
Let $\delta = \min(\delta_1, \delta_2)$. Then, if $0 < |x - c| < \delta$, both of the following inequalities hold:
$|f(x) - L_1| < \epsilon/2$
$|f(x) - L_2| < \epsilon/2$
This is because if $|x - c|$ is less than $\delta$, it is also less than or equal to both $\delta_1$ and $\delta_2$.
4. Use the Triangle Inequality:
Consider the absolute difference between $L_1$ and $L_2$:
$$|L_1 - L_2| = |L_1 - f(x) + f(x) - L_2|$$
Using the triangle inequality $|a + b| \le |a| + |b|$, we get:
$$|L_1 - L_2| \le |L_1 - f(x)| + |f(x) - L_2|$$
We can rewrite $|L_1 - f(x)|$ as $|f(x) - L_1|$, so:
$$|L_1 - L_2| \le |f(x) - L_1| + |f(x) - L_2|$$
5. Substitute the Inequalities:
From step 3, we know that for $0 < |x - c| < \delta$:
$|f(x) - L_1| < \epsilon/2$
$|f(x) - L_2| < \epsilon/2$
Substituting these into the inequality from step 4:
$$|L_1 - L_2| < \epsilon/2 + \epsilon/2$$
$$|L_1 - L_2| < \epsilon$$
6. Conclude that $L_1 = L_2$:
The inequality $|L_1 - L_2| < \epsilon$ holds for any $\epsilon > 0$. The only non-negative number that is less than every positive number is zero. Therefore, we must have:
$$|L_1 - L_2| = 0$$
This implies:
$$L_1 - L_2 = 0$$
$$L_1 = L_2$$
