Question:

Step 1: Substitute $x = 0$ into the given limit condition. We are given that $\displaystyle\lim_{t \to x} \dfrac{\sec x \cdot f(t) - f(x) \sec t}{t - 1} = \sec^2 x$ and $f(0) = 1$. Setting $x = 0$: $$\lim_{t \to 0} \dfrac{\sec 0 \cdot f(t) - f(0) \sec t}{t - 1} = \sec^2 0$$ Step 2: Simplify using $\sec 0 = 1$ and $f(0) = 1$. Since $\sec 0 = 1$ and $f(0) = 1$: $$\lim_{t \to 0} \dfrac{1 \cdot f(t) - 1 \cdot \sec t}{t - 1} = 1$$ $$\lim_{t \to 0} \dfrac{f(t) - \sec t}{t - 1} = 1$$ Step 3: Use the fact that $\sec t \to 1$ as $t \to 0$. As $t \to 0$, we have $\sec t \to \sec 0 = 1$. Therefore, $f(t) - \sec t \to f(0) - 1 = 1 - 1 = 0$. This means: $$\lim_{t \to 0} \dfrac{f(t) - 1}{t - 1} = 1$$ Step 4: Relate the given limit to $f'(0)$. We need to find $f'(0) = \displaystyle\lim_{t \to 0} \dfrac{f(t) - f(0)}{t - 0} = \lim_{t \to 0} \dfrac{f(t) - 1}{t}$. From Step 3, we have: $$\lim_{t \to 0} \dfrac{f(t) - 1}{t - 1} = 1$$ Step 5: Extract $f'(0)$ using algebraic manipulation. We can rewrite the limit from Step 4 as: $$\lim_{t \to 0} \dfrac{f(t) - 1}{t - 1} = \lim_{t \to 0} \dfrac{f(t) - 1}{t} \cdot \dfrac{t}{t - 1} = 1$$ As $t \to 0$: $$\dfrac{t}{t - 1} \to \dfrac{0}{0 - 1} = 0 - (-1) = -1$$ Therefore: $$\lim_{t \to 0} \dfrac{f(t) - 1}{t} \cdot (-1) = 1$$ $$\lim_{t \to 0} \dfrac{f(t) - 1}{t} = -1$$ Thus, $f'(0) = -1$. Step 6: Calculate $\dfrac{f(0)}{f'(0)}$. Given $f(0) = 1$ and $f'(0) = -1$: $$\dfrac{f(0)}{f'(0)} = \dfrac{1}{-1} = -1$$ **Final Answer:** The value of $\dfrac{f(0)}{f'(0)} = -1$, which corresponds to **Option 1**.