1. **Problem Statement:** Prove that if there exists a number $M$ such that $$\left|\frac{f(x) - L}{x - a}\right| \leq M, \quad \forall x \neq a,$$ then $$\lim_{x \to a} f(x) = L.$$ 2. **Understanding the problem:** We want to show that $f(x)$ approaches $L$ as $x$ approaches $a$. The given inequality bounds the difference quotient $\frac{f(x)-L}{x-a}$ by $M$. 3. **Rewrite the inequality:** Multiply both sides by $|x - a|$ (which is positive for $x \neq a$): $$|f(x) - L| = \left|\frac{f(x) - L}{x - a}\right| \cdot |x - a| \leq M |x - a|.$$ 4. **Use the definition of limit:** To prove $\lim_{x \to a} f(x) = L$, for every $\epsilon > 0$, we must find $\delta > 0$ such that if $0 < |x - a| < \delta$, then $$|f(x) - L| < \epsilon.$$ 5. **Choose $\delta$:** From step 3, we have $$|f(x) - L| \leq M |x - a|.$$ To make $|f(x) - L| < \epsilon$, it suffices to have $$M |x - a| < \epsilon \implies |x - a| < \frac{\epsilon}{M}.$$ 6. **Conclusion:** Set $$\delta = \frac{\epsilon}{M}.$$ Then for all $x$ with $0 < |x - a| < \delta$, we get $$|f(x) - L| \leq M |x - a| < M \cdot \frac{\epsilon}{M} = \epsilon,$$ which proves $$\lim_{x \to a} f(x) = L.$$