Step 1: Check the indeterminate form:
Let $u = \pi/2 - x$, so $\cos x = \sin u$ and as $x \to \pi/2$, $u \to 0$. The limit becomes $\lim_{u \to 0} \dfrac{\sin(\sin u) - \sin u}{u}$. At $u = 0$, the numerator is $\sin(\sin 0) - \sin 0 = \sin(0) - 0 = 0$ and the denominator is also $0$, so this is a $0/0$ indeterminate form and L'Hopital's rule can be applied directly.
Step 2: Differentiate numerator and denominator with respect to u:
Let $f(u) = \sin(\sin u) - \sin u$ and $g(u) = u$. Using the chain rule, $f'(u) = \cos(\sin u) \cdot \cos u - \cos u = \cos u \left[ \cos(\sin u) - 1 \right]$, and $g'(u) = 1$.
Step 3: Evaluate the derivatives at u = 0:
At $u = 0$: $\cos u = \cos 0 = 1$, and $\sin u = \sin 0 = 0$, so $\cos(\sin u) = \cos 0 = 1$. Therefore $f'(0) = 1 \times (1 - 1) = 1 \times 0 = 0$, and $g'(0) = 1$.
Step 4: Apply L'Hopital's rule:
By L'Hopital's rule, $\lim_{u \to 0} \dfrac{f(u)}{g(u)} = \dfrac{f'(0)}{g'(0)} = \dfrac{0}{1} = 0$. Since the derivative of the denominator is nonzero at $u=0$, a single application of the rule is sufficient, and this confirms the same result obtained from the Taylor series expansion method.
Final Answer:
\[\boxed{0}\]