Limit x=>π/2ln(1+sinx/(1-sinx))..Please help me with this question.urgent guys

To find the limit as \( x \) approaches \( \frac{\pi}{2} \) of the expression \( \ln\left( \frac{1 + \sin x}{1 - \sin x} \right) \), we can start by analyzing the behavior of the sine function and the logarithm as \( x \) gets close to \( \frac{\pi}{2} \).

Understanding the Components

First, let's evaluate \( \sin x \) as \( x \) approaches \( \frac{\pi}{2} \). We know that:

• \( \sin\left(\frac{\pi}{2}\right) = 1 \)

This means that as \( x \) approaches \( \frac{\pi}{2} \), \( \sin x \) approaches 1. Now, substituting this into our expression gives:

• \( 1 + \sin x \) approaches \( 1 + 1 = 2 \)
• \( 1 - \sin x \) approaches \( 1 - 1 = 0 \)

Evaluating the Limit

Now, substituting these values into the logarithmic expression, we have:

As \( x \) approaches \( \frac{\pi}{2} \), the expression becomes:

\( \ln\left( \frac{2}{0} \right) \)

Since \( \frac{2}{0} \) tends toward infinity, we can conclude that:

\( \ln\left( \frac{2}{0} \right) \) approaches \( \ln(\infty) \), which is \( +\infty \).

Final Result

Thus, the limit can be expressed as:

\( \lim_{x \to \frac{\pi}{2}} \ln\left( \frac{1 + \sin x}{1 - \sin x} \right) = +\infty \).

In summary, as \( x \) approaches \( \frac{\pi}{2} \), the logarithmic expression diverges to positive infinity due to the denominator approaching zero while the numerator remains finite. This behavior is typical in limits involving logarithmic functions where the argument approaches a form that leads to infinity.