Question: top-right: $\lim_{x \to 0} \frac{e^{7x}-1-7x}{x^2}$
1. **State the problem:**
Find the limit $$\lim_{x \to 0} \frac{e^{7x}-1-7x}{x^2}$$.
2. **Recall the formula and rules:**
The exponential function can be expanded using its Taylor series around 0:
$$e^{7x} = 1 + 7x + \frac{(7x)^2}{2!} + \frac{(7x)^3}{3!} + \cdots$$
3. **Apply the Taylor expansion:**
Substitute the expansion into the numerator:
$$e^{7x} - 1 - 7x = \left(1 + 7x + \frac{49x^2}{2} + \cdots \right) - 1 - 7x = \frac{49x^2}{2} + \cdots$$
4. **Rewrite the limit using the expansion:**
$$\lim_{x \to 0} \frac{\frac{49x^2}{2} + \cdots}{x^2}$$
5. **Simplify the fraction:**
$$\lim_{x \to 0} \frac{49x^2/2 + \cdots}{x^2} = \lim_{x \to 0} \left( \frac{49}{2} + \cdots \right)$$
6. **Evaluate the limit:**
As $x \to 0$, higher order terms vanish, so the limit is:
$$\boxed{\frac{49}{2}}$$
**Final answer:** $$\lim_{x \to 0} \frac{e^{7x}-1-7x}{x^2} = \frac{49}{2}$$.