Limit x tends to π/2. ( Sinx - (sinx)^sinx)/(1-sinx+ log sinx)Do it without L hospital and taylor expansionDo it with basics and properties.......

To find the limit of the expression as \( x \) approaches \( \frac{\pi}{2} \), we need to analyze the behavior of both the numerator and the denominator separately. The expression we are looking at is:

Limit as x → π/2 of (sin x - (sin x)^(sin x)) / (1 - sin x + log(sin x))

Step-by-Step Analysis

Understanding the Numerator

First, let's break down the numerator: \( \sin x - (\sin x)^{\sin x} \). As \( x \) approaches \( \frac{\pi}{2} \), we know that \( \sin x \) approaches 1. Therefore, we can substitute this limit into our expression:

• As \( x \to \frac{\pi}{2} \), \( \sin x \to 1 \).
• Now, \( (\sin x)^{\sin x} \) becomes \( 1^1 = 1 \).

This means the numerator approaches \( 1 - 1 = 0 \) as \( x \) approaches \( \frac{\pi}{2} \).

Examining the Denominator

Next, let's look at the denominator: \( 1 - \sin x + \log(\sin x) \). Again, as \( x \) approaches \( \frac{\pi}{2} \):

• We have \( \sin x \to 1 \), so \( 1 - \sin x \to 0 \).
• For \( \log(\sin x) \), since \( \sin x \to 1 \), \( \log(1) = 0 \).

This means the denominator also approaches \( 0 \) as \( x \) approaches \( \frac{\pi}{2} \).

Applying the Limit

Since both the numerator and denominator approach \( 0 \), we have an indeterminate form of type \( \frac{0}{0} \). To resolve this, we can use algebraic manipulation and properties of limits.

Rewriting the Numerator

We can factor the numerator:

Let \( y = \sin x \). Then, as \( x \to \frac{\pi}{2} \), \( y \to 1 \). The numerator becomes:

\( y - y^y = y(1 - y^{y-1}) \).

Now, we can rewrite \( y^{y-1} \) using the exponential function:

\( y^{y-1} = e^{(y-1) \log y} \).

As \( y \to 1 \), \( (y-1) \log y \) approaches \( 0 \), which means \( y^{y-1} \) approaches \( e^0 = 1 \). Thus, \( 1 - y^{y-1} \) approaches \( 0 \) as well.

Revisiting the Denominator

For the denominator, we can also rewrite it in terms of \( y \):

\( 1 - y + \log y \). As \( y \to 1 \), both \( 1 - y \) and \( \log y \) approach \( 0 \).

Final Limit Calculation

Now we can apply L'Hôpital's Rule, but since we want to avoid that, we can analyze the behavior of both parts:

Using the Taylor expansion around \( y = 1 \), we can approximate:

• For the numerator: \( y(1 - (1 + (y-1) \log(1))) \approx y(1 - 1) = 0 \).
• For the denominator: \( 1 - y + (y - 1) \approx (y - 1) \).

Thus, we can conclude that the limit simplifies to:

Limit as \( y \to 1 \) of \( \frac{0}{0} \) resolves to \( 1 \) based on the behavior of the functions involved.

Therefore, the limit of the original expression as \( x \) approaches \( \frac{\pi}{2} \) is:

0