Question

Evaluate: \[ \lim_{x\to \frac{\pi}{2}} \left( \frac{1-\sin x}{\cos x} \right) \]

• A. \(\dfrac12\)
• B. \(-1\)
• C. \(0\)
• D. \(1\)

Hint:

For limits involving \(1-\sin x\), multiply by the conjugate \(1+\sin x\). It usually converts the expression into a simpler trigonometric ratio.

Answer 1

The Correct Option is C

Concept: When trigonometric limits produce indeterminate forms, rationalization is often the simplest method. The identity used here is: \[ 1-\sin x = \frac{(1-\sin x)(1+\sin x)}{1+\sin x} \] which simplifies the expression significantly.

Step 1: Observe the indeterminate form. We are given: \[ \lim_{x\to \frac{\pi}{2}} \frac{1-\sin x}{\cos x} \] Substituting directly: \[ \sin\frac{\pi}{2}=1, \qquad \cos\frac{\pi}{2}=0 \] Hence, \[ \frac{1-1}{0} = \frac00 \] which is an indeterminate form.

Step 2: Rationalize the numerator. Multiply numerator and denominator by \(1+\sin x\): \[ \frac{1-\sin x}{\cos x} \cdot \frac{1+\sin x}{1+\sin x} \] \[ = \frac{1-\sin^2x}{\cos x(1+\sin x)} \] Using the identity: \[ 1-\sin^2x=\cos^2x \] we get: \[ = \frac{\cos^2x}{\cos x(1+\sin x)} \] Cancel one \(\cos x\): \[ = \frac{\cos x}{1+\sin x} \]

Step 3: Evaluate the limit. Now substitute \(x\to \frac{\pi}{2}\): \[ \cos\frac{\pi}{2}=0 \] and \[ 1+\sin\frac{\pi}{2}=1+1=2 \] Therefore, \[ \lim_{x\to \frac{\pi}{2}} \frac{\cos x}{1+\sin x} = \frac02 \] \[ =0 \] Hence, \[ \boxed{0} \]