Question:

Step 1: Identify the need to use Taylor series expansions. To evaluate the limit of the product of three expressions as $x \to 0$, we need to find the Taylor series expansion of each component around $x = 0$. Step 2: Find the Taylor series expansion of $1 - \cos x$. Using the standard Taylor series for cosine: $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$ Therefore: $$1 - \cos x = \frac{x^2}{2} - \frac{x^4}{24} + \cdots \approx \frac{x^2}{2}$$ Step 3: Find the Taylor series expansion of $\sin x - x$. Using the standard Taylor series for sine: $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ Therefore: $$\sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - \cdots \approx -\frac{x^3}{6}$$ Step 4: Find the Taylor series expansion of $e^x + e^{-x} - 2$. Using the standard Taylor series for exponentials: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ $$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots$$ Adding these: $$e^x + e^{-x} = 2 + \frac{2x^2}{2!} + \frac{2x^4}{4!} + \cdots$$ Therefore: $$e^x + e^{-x} - 2 = x^2 + \frac{x^4}{12} + \cdots \approx x^2$$ Step 5: Multiply the three expansions together. The product of the three expressions is: $$(1 - \cos x)(\sin x - x)(e^x + e^{-x} - 2) \approx \left(\frac{x^2}{2}\right)\left(-\frac{x^3}{6}\right)(x^2)$$ $$= -\frac{x^2 \cdot x^3 \cdot x^2}{2 \cdot 6} = -\frac{x^7}{12}$$ Step 6: Determine the value of $n$ for which the limit is finite and non-zero. For the limit to be finite and non-zero: $$\lim_{x \to 0} \frac{-\frac{x^7}{12}}{x^n} = \lim_{x \to 0} -\frac{x^{7-n}}{12}$$ This limit is finite and non-zero only when the exponent of $x$ equals zero, which means: $$7 - n = 0 \implies n = 7$$ When $n = 7$, the limit equals $-\frac{1}{12}$, which is finite and non-zero. **Final Answer:** The minimum integral value of $n$ is $\boxed{7}$, which corresponds to **Option 3**.