Question:

Step 1: Set up the limit expression and rewrite the base in a more convenient form. We need to find: $$L = \lim_{x \to \infty} x\left(\left(\frac{x}{x+1}\right)^x - \frac{1}{e}\right)$$ First, rewrite the base as: $$\frac{x}{x+1} = 1 - \frac{1}{x+1}$$ Step 2: Expand the logarithm of the expression using Taylor series. To analyze $\left(\frac{x}{x+1}\right)^x$, we compute its logarithm: $$\ln\left(\left(\frac{x}{x+1}\right)^x\right) = x\ln\left(1-\frac{1}{x+1}\right)$$ Using the Taylor expansion $\ln(1-u) = -u - \frac{u^2}{2} - \frac{u^3}{3} - \cdots$ with $u = \frac{1}{x+1}$: $$x\ln\left(1-\frac{1}{x+1}\right) = x\left(-\frac{1}{x+1} - \frac{1}{2(x+1)^2} - \frac{1}{3(x+1)^3} - \cdots\right)$$ Step 3: Simplify the logarithmic expansion. Let $u = \frac{1}{x+1}$, so $x = \frac{1}{u} - 1$. Then: $$x\ln(1-u) = \left(\frac{1}{u}-1\right)\ln(1-u) = \left(\frac{1}{u}-1\right)\left(-u - \frac{u^2}{2} - \frac{u^3}{3} - \cdots\right)$$ Expanding: $$= -\left(\frac{1}{u}-1\right)u\left(1 + \frac{u}{2} + \frac{u^2}{3} + \cdots\right)$$ $$= -(1-u)\left(1 + \frac{u}{2} + \frac{u^2}{3} + \cdots\right)$$ $$= -\left(1 + \frac{u}{2} + \frac{u^2}{3} - u - \frac{u^2}{2} + \cdots\right)$$ $$= -1 + \frac{u}{2} + \frac{u^2}{6} + O(u^3)$$ Step 4: Express the original expression using the exponential form. Since $\ln\left(\left(\frac{x}{x+1}\right)^x\right) = -1 + \frac{u}{2} + O(u^2)$, we have: $$\left(\frac{x}{x+1}\right)^x = e^{-1+\frac{u}{2}+O(u^2)} = \frac{1}{e} \cdot e^{\frac{u}{2}+O(u^2)}$$ Using the expansion $e^v \approx 1 + v$ for small $v$: $$\left(\frac{x}{x+1}\right)^x = \frac{1}{e}\left(1 + \frac{u}{2} + O(u^2)\right)$$ Step 5: Compute the difference and multiply by $x$. $$\left(\frac{x}{x+1}\right)^x - \frac{1}{e} = \frac{1}{e}\left(1 + \frac{u}{2} + O(u^2)\right) - \frac{1}{e} = \frac{1}{e} \cdot \frac{u}{2} + O(u^2)$$ Therefore: $$x\left(\left(\frac{x}{x+1}\right)^x - \frac{1}{e}\right) = x \cdot \frac{1}{e} \cdot \frac{u}{2} + O(xu^2)$$ Step 6: Evaluate the limit. Substituting $u = \frac{1}{x+1}$: $$x \cdot \frac{1}{e} \cdot \frac{u}{2} = \frac{x}{e} \cdot \frac{1}{2(x+1)} = \frac{1}{e} \cdot \frac{x}{2(x+1)}$$ As $x \to \infty$: $$\frac{x}{2(x+1)} = \frac{1}{2\left(1+\frac{1}{x}\right)} \to \frac{1}{2}$$ Therefore: $$L = \frac{1}{e} \cdot \frac{1}{2} = \frac{1}{2e}$$ **Final Answer:** The limit equals $\boxed{\frac{1}{2e}}$, which corresponds to **Option 2**.