1. **State the problem:** Evaluate the limit $$\lim_{x \to 0} \frac{e^x - x - 1}{5x^2}.$$\n\n2. **Recall the formula and rules:** When direct substitution results in an indeterminate form like $$\frac{0}{0},$$ we can use L'Hôpital's rule, which states:\n$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ if the latter limit exists.\n\n3. **Check direct substitution:**\n$$e^0 - 0 - 1 = 1 - 0 - 1 = 0,$$ and denominator $$5 \cdot 0^2 = 0,$$ so the limit is of the form $$\frac{0}{0}$$, suitable for L'Hôpital's rule.\n\n4. **Apply L'Hôpital's rule (first derivative):**\n$$f(x) = e^x - x - 1 \implies f'(x) = e^x - 1,$$\n$$g(x) = 5x^2 \implies g'(x) = 10x.$$\nSo the limit becomes:\n$$\lim_{x \to 0} \frac{e^x - 1}{10x}.$$\n\n5. **Check direct substitution again:**\n$$e^0 - 1 = 0,$$ denominator $$10 \cdot 0 = 0,$$ still $$\frac{0}{0}$$, apply L'Hôpital's rule again.\n\n6. **Apply L'Hôpital's rule (second derivative):**\n$$f''(x) = e^x,$$\n$$g''(x) = 10.$$\nLimit becomes:\n$$\lim_{x \to 0} \frac{e^x}{10} = \frac{e^0}{10} = \frac{1}{10}.$$\n\n7. **Final answer:**\n$$\boxed{\frac{1}{10}}.$$