Question:

Step 1: Define an antiderivative function. Let us define $F(x) = \int e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right) dx$. By the Fundamental Theorem of Calculus, the derivative of this function is: $$F'(x) = e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right)$$ Step 2: Rewrite the integral using the antiderivative. Using the Fundamental Theorem of Calculus, we can express the definite integral as a difference of antiderivative values: $$\int_{1}^{1+2h} e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right) dx = F(1+2h) - F(1)$$ Therefore, the given limit becomes: $$\lim_{h \to 0} \frac{1}{h} \left[ F(1+2h) - F(1) \right]$$ Step 3: Apply the definition of derivative to evaluate the limit. We recognize that this limit has the form of a derivative. Specifically, we can rewrite it as: $$\lim_{h \to 0} \frac{1}{h} \left[ F(1+2h) - F(1) \right] = \lim_{h \to 0} \frac{F(1+2h) - F(1)}{2h} \cdot 2$$ By the definition of the derivative, $\displaystyle\lim_{h \to 0} \frac{F(1+2h) - F(1)}{2h} = F'(1)$. Therefore: $$\lim_{h \to 0} \frac{1}{h} \left[ F(1+2h) - F(1) \right] = 2F'(1)$$ Step 4: Evaluate $F'(1)$. Substituting $x = 1$ into the derivative $F'(x) = e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right)$: $$F'(1) = e^{\sqrt{1}} \sin\left(\frac{\pi \cdot 1}{3}\right) = e^{1} \sin\left(\frac{\pi}{3}\right) = e\sin\left(\frac{\pi}{3}\right)$$ Step 5: Calculate the final answer. Substituting $F'(1)$ back into our expression from Step 3: $$\lim_{h \to 0} \frac{1}{h} \int_{1}^{1+2h} e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right) dx = 2F'(1) = 2e\sin\left(\frac{\pi}{3}\right)$$ **Final Answer:** The value of the given limit is $\boxed{2e\sin\dfrac{\pi}{3}}$, which corresponds to **Option 4**.