Question:

Step 1: Simplify the numerator of the fraction in $f(x)$. The numerator is $x\ln x - \ln x = \ln x(x - 1)$. At $x = 1$, this equals $\ln(1) \cdot (1-1) = 0 \cdot 0 = 0$. Step 2: Simplify the denominator of the fraction in $f(x)$. The denominator is: $$9x^2 - 2e^x x - 9x + 2e^x$$ Rearranging by grouping: $$9x^2 - 9x - 2e^x x + 2e^x = 9x(x-1) - 2e^x(x-1) = (x-1)(9x - 2e^x)$$ At $x = 1$, this equals $(1-1)(9 - 2e) = 0$. Step 3: Recognize the indeterminate form and cancel common factors. Since both numerator and denominator approach $0$ as $x \to 1$, we have a $\dfrac{0}{0}$ indeterminate form. We can cancel the common factor $(x-1)$: $$f(x) - 2 = \dfrac{\ln x(x-1)}{(x-1)(9x - 2e^x)} = \dfrac{\ln x}{9x - 2e^x}$$ Step 4: Evaluate $f(x)$ at $x = 1$. $$\lim_{x \to 1} \left(f(x) - 2\right) = \lim_{x \to 1} \dfrac{\ln x}{9x - 2e^x} = \dfrac{\ln 1}{9(1) - 2e^1} = \dfrac{0}{9 - 2e} = 0$$ Therefore: $$\lim_{x \to 1} f(x) = 2 + 0 = 2$$ Step 5: Evaluate $g(x)$ at $x = 1$. $$g(1) = \sin^2\left(\dfrac{\pi(1)^2}{2}\right) = \sin^2\left(\dfrac{\pi}{2}\right) = 1^2 = 1$$ Step 6: Calculate the limit of the ratio. Since $\lim_{x \to 1} f(x) = 2$ and $\lim_{x \to 1} g(x) = 1$, we have: $$\lim_{x \to 1} \dfrac{f(x)}{g(x)} = \dfrac{2}{1} = 2$$ **Final Answer:** The value of $\lim_{x \to 1} \dfrac{f(x)}{g(x)} = 2$, which corresponds to **Option 1**.