1. **Problem Statement:** Prove that if the limit of a function $f(x)$ as $x$ approaches $x_0$ exists, then this limit is unique. 2. **Definition of Limit:** The limit of $f(x)$ as $x$ approaches $x_0$ is $l$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0 < |x - x_0| < \delta$, it follows that $|f(x) - l| < \epsilon$. 3. **Assume for Contradiction:** Suppose the limit exists but is not unique. That means there are two limits $l_1$ and $l_2$ with $l_1 \neq l_2$ such that: $$\lim_{x \to x_0} f(x) = l_1 \quad \text{and} \quad \lim_{x \to x_0} f(x) = l_2$$ 4. **Apply the Definition to Both Limits:** For $\epsilon = \frac{|l_1 - l_2|}{3} > 0$, there exist $\delta_1, \delta_2 > 0$ such that: - If $0 < |x - x_0| < \delta_1$, then $|f(x) - l_1| < \epsilon$ - If $0 < |x - x_0| < \delta_2$, then $|f(x) - l_2| < \epsilon$ 5. **Choose $\delta = \min(\delta_1, \delta_2)$:** For $0 < |x - x_0| < \delta$, both inequalities hold simultaneously. 6. **Use Triangle Inequality:** $$|l_1 - l_2| = |l_1 - f(x) + f(x) - l_2| \leq |l_1 - f(x)| + |f(x) - l_2| < \epsilon + \epsilon = 2\epsilon$$ 7. **Contradiction:** But $2\epsilon = 2 \times \frac{|l_1 - l_2|}{3} = \frac{2}{3}|l_1 - l_2| < |l_1 - l_2|$, which contradicts the inequality above. 8. **Conclusion:** Our assumption that two distinct limits exist is false. Therefore, the limit of $f(x)$ as $x$ approaches $x_0$ is unique if it exists.