1. **Problem statement:** Find the limit $$\lim_{x \to a} \frac{x^{1/m} - a^{1/m}}{x - a}$$ without using L'Hospital's rule or differentiation. 2. **Recall the formula for difference of powers:** For any real numbers $u$ and $v$, and integer $m$, we have $$u^m - v^m = (u - v)(u^{m-1} + u^{m-2}v + \cdots + uv^{m-2} + v^{m-1}).$$ 3. **Apply substitution:** Let $$u = x^{1/m}, \quad v = a^{1/m}.$$ Then $$u^m = x, \quad v^m = a.$$ So, $$x - a = u^m - v^m = (u - v)(u^{m-1} + u^{m-2}v + \cdots + uv^{m-2} + v^{m-1}).$$ 4. **Rewrite the original expression:** $$\frac{x^{1/m} - a^{1/m}}{x - a} = \frac{u - v}{(u - v)(u^{m-1} + u^{m-2}v + \cdots + v^{m-1})} = \frac{1}{u^{m-1} + u^{m-2}v + \cdots + v^{m-1}}.$$ 5. **Evaluate the limit as $x \to a$:** Then $u \to v$, so $$\lim_{x \to a} \frac{x^{1/m} - a^{1/m}}{x - a} = \frac{1}{m v^{m-1}} = \frac{1}{m (a^{1/m})^{m-1}} = \frac{1}{m a^{(m-1)/m}}.$$ **Final answer:** $$\boxed{\lim_{x \to a} \frac{x^{1/m} - a^{1/m}}{x - a} = \frac{1}{m a^{\frac{m-1}{m}}}}.$$