1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{(1 + mx)^n - (1 + nx)^m}{x^2}$$ without using L'Hôpital's rule. 2. **Recall the binomial expansion for small $x$:** For any real number $a$ and integer $k$, when $x$ is close to 0, $$ (1 + ax)^k \approx 1 + kax + \frac{k(k-1)}{2}a^2x^2 $$ This approximation includes terms up to $x^2$ because higher powers vanish faster as $x \to 0$. 3. **Apply the expansion to each term:** - Expand $(1 + mx)^n$: $$ (1 + mx)^n \approx 1 + nmx + \frac{n(n-1)}{2}m^2x^2 $$ - Expand $(1 + nx)^m$: $$ (1 + nx)^m \approx 1 + mnx + \frac{m(m-1)}{2}n^2x^2 $$ 4. **Substitute expansions into the limit expression:** $$ \frac{(1 + mx)^n - (1 + nx)^m}{x^2} \approx \frac{\left(1 + nmx + \frac{n(n-1)}{2}m^2x^2\right) - \left(1 + mnx + \frac{m(m-1)}{2}n^2x^2\right)}{x^2} $$ 5. **Simplify numerator:** - The constant terms $1 - 1 = 0$ - The linear terms $nmx - mnx = 0$ (since $n m x = m n x$) - The quadratic terms: $$ \frac{n(n-1)}{2}m^2x^2 - \frac{m(m-1)}{2}n^2x^2 = x^2 \left( \frac{n(n-1)}{2}m^2 - \frac{m(m-1)}{2}n^2 \right) $$ 6. **Divide numerator by $x^2$ and take the limit as $x \to 0$:** $$ \lim_{x \to 0} \frac{(1 + mx)^n - (1 + nx)^m}{x^2} = \frac{n(n-1)}{2}m^2 - \frac{m(m-1)}{2}n^2 $$ 7. **Final answer:** $$ \boxed{\frac{n(n-1)}{2}m^2 - \frac{m(m-1)}{2}n^2} $$ This result shows the limit depends on $m$ and $n$ through these quadratic expressions.