Evaluate the following limits :
`Lim_(x to pi/2) (cos^(2)x)/(1-sinx ) `

Answer

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x} \]

we will follow these steps:

### Step 1: Rewrite \(\cos^2 x\)
Using the Pythagorean identity, we know that

\[ \cos^2 x = 1 - \sin^2 x. \]

### Step 2: Substitute into the limit
Substituting this identity into the limit gives us:

\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^2 x}{1 - \sin x}. \]

### Step 3: Factor the numerator
The expression \(1 - \sin^2 x\) can be factored as a difference of squares:

\[ 1 - \sin^2 x = (1 - \sin x)(1 + \sin x). \]

### Step 4: Substitute the factored form
Now, substituting this back into the limit, we have:

\[ \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)(1 + \sin x)}{1 - \sin x}. \]

### Step 5: Cancel common terms
Since \(1 - \sin x\) is in both the numerator and the denominator, we can cancel it (as long as \(x \neq \frac{\pi}{2}\)):

\[ \lim_{x \to \frac{\pi}{2}} (1 + \sin x). \]

### Step 6: Evaluate the limit
Now we can directly substitute \(x = \frac{\pi}{2}\):

\[ 1 + \sin\left(\frac{\pi}{2}\right) = 1 + 1 = 2. \]

### Final Answer
Thus, the limit is

\[ \boxed{2}. \]

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