Exponential Limit F7A249

1. **Problem:** Find the limit as $x$ approaches $0$ of $$\frac{e^{4x}-1}{x}$$. 2. **Use a standard limit formula:** A very important limit is $$\lim_{u\to 0}\frac{e^u-1}{u}=1.$$ This works because the exponential function changes smoothly near $0$. 3. **Rewrite the expression to match the formula:** Let $u=4x$. Then as $x\to 0$, we also have $u\to 0$. Substitute $u=4x$: $$\frac{e^{4x}-1}{x}=\frac{e^u-1}{u}\cdot\frac{u}{x}.$$ Since $u=4x$, we get $$\frac{u}{x}=\frac{4x}{x}=4.$$ So the expression becomes $$\frac{e^{4x}-1}{x}=4\cdot\frac{e^u-1}{u}.$$ 4. **Take the limit:** $$\lim_{x\to 0}\frac{e^{4x}-1}{x}=4\cdot\lim_{u\to 0}\frac{e^u-1}{u}=4\cdot 1=4.$$ 5. **Final answer:** $$\boxed{4}$$