Evaluate the following limits :
`lim_(x to pi/2)(cotx-cosx)/(cos^(3)x)`

Answer

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{2}} \frac{\cot x - \cos x}{\cos^3 x}, \]

we start by substituting \( x = \frac{\pi}{2} \) into the expression.

1. **Substitution**:
\[ \cot\left(\frac{\pi}{2}\right) = 0, \quad \cos\left(\frac{\pi}{2}\right) = 0, \quad \cos^3\left(\frac{\pi}{2}\right) = 0. \]
This gives us the indeterminate form \( \frac{0 - 0}{0} = \frac{0}{0} \).

2. **Rewrite the expression**:
We can rewrite \( \cot x \) in terms of sine and cosine:
\[ \cot x = \frac{\cos x}{\sin x}. \]
Thus, we can express the limit as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\frac{\cos x}{\sin x} - \cos x}{\cos^3 x}. \]

3. **Combine the terms**:
The expression in the numerator becomes:
\[ \frac{\cos x - \cos x \sin x}{\sin x} = \frac{\cos x(1 - \sin x)}{\sin x}. \]
Therefore, the limit can be rewritten as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x (1 - \sin x)}{\sin x \cos^3 x} = \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\sin x \cos^2 x}. \]

4. **Simplifying further**:
Now, we can factor \( 1 - \sin x \) using the identity \( 1 - \sin^2 x = \cos^2 x \):
\[ 1 - \sin x = (1 - \sin x)(1 + \sin x) / (1 + \sin x) = \frac{(1 - \sin^2 x)}{1 + \sin x} = \frac{\cos^2 x}{1 + \sin x}. \]
Thus, the limit becomes:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{\sin x \cos^2 x (1 + \sin x)}. \]

5. **Canceling terms**:
The \( \cos^2 x \) in the numerator and denominator cancels out:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1}{\sin x (1 + \sin x)}. \]

6. **Substituting the limit**:
Now substituting \( x = \frac{\pi}{2} \):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{and} \quad 1 + \sin\left(\frac{\pi}{2}\right) = 2. \]
Therefore, we have:
\[ \frac{1}{1 \cdot 2} = \frac{1}{2}. \]

Thus, the final result is:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cot x - \cos x}{\cos^3 x} = \frac{1}{2}. \]