Question

The solution of the given expression is _______. \[ \lim_{x \to \pi/2} \frac{\sin(\cos x) - \cos x}{\pi/2 - x} \]

Hint:

Substitute u = pi/2 - x so that cos x becomes sin u, then use the small angle Taylor expansion of sine.

Answer 1

Correct Answer: 0

Step 1: Set up the substitution:
Let \( u = \pi/2 - x \), so that \( x = \pi/2 - u \). As \( x \to \pi/2 \), we have \( u \to 0 \). Also, \( \cos x = \cos(\pi/2 - u) = \sin u \).
Step 2: Rewrite the given limit in terms of u:
The original expression \[ \lim_{x \to \pi/2} \frac{\sin(\cos x) - \cos x}{\pi/2 - x} \] becomes \[ \lim_{u \to 0} \frac{\sin(\sin u) - \sin u}{u} \]
Step 3: Expand sin(sin u) using the Taylor series:
For small values of \( t \), \( \sin t = t - \dfrac{t^3}{6} + \dfrac{t^5}{120} - \cdots \). Taking \( t = \sin u \), we get \[ \sin(\sin u) = \sin u - \frac{(\sin u)^3}{6} + \frac{(\sin u)^5}{120} - \cdots \]
Step 4: Substitute back and simplify the numerator:
\[ \sin(\sin u) - \sin u = -\frac{(\sin u)^3}{6} + \frac{(\sin u)^5}{120} - \cdots \] So the expression becomes \[ \frac{-\dfrac{(\sin u)^3}{6} + \cdots}{u} = -\frac{1}{6}\left(\frac{\sin u}{u}\right)(\sin u)^2 + \cdots \]
Step 5: Take the limit as u tends to 0:
As \( u \to 0 \), \( \dfrac{\sin u}{u} \to 1 \) and \( (\sin u)^2 \to 0 \). Since the leading surviving term is of order \( u^2 \), because sin u itself is of order u, the whole expression tends to 0. All higher order terms vanish even faster than this leading term.
Final Answer:
\[ \boxed{0} \]