Question:

Step 1: Identify the function and its behavior near zero. We need to find $\displaystyle\lim_{x\to 0}\left\lfloor (1-e^x)\frac{\sin x}{|x|}\right\rfloor$, where $\lfloor \cdot \rfloor$ denotes the greatest integer function. Let us define: $$g(x) = (1-e^x)\frac{\sin x}{|x|}$$ We will analyze the behavior of $g(x)$ as $x$ approaches $0$ from both sides. Step 2: Analyze the right-hand limit as $x \to 0^+$. For $x > 0$, we have $|x| = x$, so: $$g(x) = (1-e^x)\frac{\sin x}{x}$$ Using standard Taylor expansions near $x = 0$: - $(1-e^x) \approx -x$ (since $e^x \approx 1 + x$) - $\frac{\sin x}{x} \approx 1$ Therefore: $$g(x) \approx (-x) \cdot 1 = -x \to 0^-$$ This means $g(x)$ approaches $0$ from below as $x \to 0^+$. Step 3: Analyze the left-hand limit as $x \to 0^-$. For $x < 0$, we have $|x| = -x$, so: $$g(x) = (1-e^x)\frac{\sin x}{-x} = -(1-e^x)\frac{\sin x}{x}$$ Using Taylor expansions: - $(1-e^x) \approx -x$ for small $x$ - $\frac{\sin x}{x} \approx 1$ Therefore: $$g(x) \approx -(-x) \cdot 1 = x$$ Since $x < 0$ in this limit, we have $g(x) \approx x \to 0^-$. This means $g(x)$ approaches $0$ from below as $x \to 0^-$. Step 4: Determine the value of the greatest integer function. From both one-sided limits, we have established that: $$g(x) \to 0^-$$ This means $g(x)$ approaches $0$ from below, so $g(x)$ takes values in the interval $(-1, 0)$ for $x$ sufficiently close to $0$ (but $x \neq 0$). For any value $y \in (-1, 0)$, the greatest integer function gives: $$\lfloor y \rfloor = -1$$ Therefore: $$\lim_{x \to 0}\left\lfloor (1-e^x)\frac{\sin x}{|x|}\right\rfloor = -1$$ The answer is **Option 2: $-1$**