Question:

<p><strong>Step 1:</strong> Rewrite the expression by separating the product:</p><p>$$\lim_{x \to 0} \left(\frac{x^2 \sin(1/x) + 2x}{x}\right) \cdot \frac{x}{(1+x)^{1/x} - e}$$</p><p><strong>Step 2:</strong> Simplify the first fraction:</p><p>$$\frac{x^2 \sin(1/x) + 2x}{x} = x\sin\left(\frac{1}{x}\right) + 2$$</p><p>Since $|\sin(1/x)| \leq 1$, by squeeze theorem: $\lim_{x \to 0} x\sin(1/x) = 0$</p><p>Therefore: $\lim_{x \to 0} \left(x\sin(1/x) + 2\right) = 2$</p><p><strong>Step 3:</strong> Evaluate the second fraction using Taylor expansion of $(1+x)^{1/x}$:</p><p>$(1+x)^{1/x} = e^{\frac{\ln(1+x)}{x}} = e^{\frac{(x - x^2/2 + x^3/3 - ...)}{x}} = e^{1 - x/2 + x^2/3 - ...}$</p><p>$$= e \cdot e^{-x/2 + x^2/3 - ...} = e\left(1 - \frac{x}{2} + O(x^2)\right)$$</p><p><strong>Step 4:</strong> Find the denominator:</p><p>$$(1+x)^{1/x} - e = e\left(-\frac{x}{2} + O(x^2)\right) = -\frac{ex}{2} + O(x^2)$$</p><p><strong>Step 5:</strong> Compute the second limit:</p><p>$$\lim_{x \to 0} \frac{x}{-\frac{ex}{2} + O(x^2)} = \lim_{x \to 0} \frac{x}{-\frac{ex}{2}} = \frac{1}{-e/2} = -\frac{2}{e}$$</p><p><strong>Step 6:</strong> Multiply the results:</p><p>$$2 \cdot \left(-\frac{2}{e}\right) = -\frac{4}{e}$$</p><p>∴ Answer: D</p>