Limit Exponential C4E455

1. **State the problem:** We want to find the limit $$\lim_{x \to 0} \frac{e^{4x} - 1}{x}.$$\n\n2. **Recall the formula and rules:** The exponential function $e^u$ can be expanded using its Taylor series around $u=0$ as $$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots.$$\nWhen $x$ approaches $0$, $4x$ also approaches $0$, so we can use this expansion for $e^{4x}$.\n\n3. **Apply the expansion:** Substitute $u = 4x$:\n$$e^{4x} = 1 + 4x + \frac{(4x)^2}{2!} + \cdots = 1 + 4x + 8x^2 + \cdots.$$\n\n4. **Rewrite the numerator:**\n$$e^{4x} - 1 = (1 + 4x + 8x^2 + \cdots) - 1 = 4x + 8x^2 + \cdots.$$\n\n5. **Form the fraction:**\n$$\frac{e^{4x} - 1}{x} = \frac{4x + 8x^2 + \cdots}{x}.$$\n\n6. **Simplify by canceling $x$:**\n$$\frac{\cancel{x}(4 + 8x + \cdots)}{\cancel{x}} = 4 + 8x + \cdots.$$\n\n7. **Take the limit as $x \to 0$:**\n$$\lim_{x \to 0} (4 + 8x + \cdots) = 4.$$\n\n**Final answer:** $$\boxed{4}.$$