Let `f: R -> R` be a differentiable function such that `f(0) = 0,f(pi/2)=3a n df^(prime)(0)=1.` If `g(x)=int_x^(pi/2)[f^(prime)(t)cos e ct-cottcos e ctf(t)]dt` for `x(0,pi/2],` then `(lim)_(x -> 0)g(x)=`

Explanation

To solve the given problem step by step, we will follow the reasoning provided in the video transcript and derive the solution systematically.

### Step 1: Understanding the Function and Given Conditions
We are given a differentiable function \( f: \mathbb{R} \to \mathbb{R} \) with the following properties:
- \( f(0) = 0 \)
- \( f\left(\frac{\pi}{2}\right) = 3 \)
- \( f'(0) = 1 \)

We need to analyze the function \( g(x) \) defined as:
\[ g(x) = \int_x^{\frac{\pi}{2}} \left[ f'(t) \csc t - \cot t \csc t f(t) \right] dt \]
for \( x \in (0, \frac{\pi}{2}] \).

### Step 2: Splitting the Integral
We can split \( g(x) \) into two separate integrals:
\[ g(x) = \int_x^{\frac{\pi}{2}} f'(t) \csc t \, dt - \int_x^{\frac{\pi}{2}} \cot t \csc t f(t) \, dt \]
Let’s denote the first integral as \( I_1 \) and the second integral as \( I_2 \):
\[ g(x) = I_1 - I_2 \]

### Step 3: Evaluating \( I_1 \) using Integration by Parts
For \( I_1 \):
\[ I_1 = \int_x^{\frac{\pi}{2}} f'(t) \csc t \, dt \]
We can use integration by parts, letting:
- \( u = \csc t \) (thus \( du = -\csc t \cot t \, dt \))
- \( dv = f'(t) \, dt \) (thus \( v = f(t) \))

Applying integration by parts:
\[ I_1 = \left[ f(t) \csc t \right]_x^{\frac{\pi}{2}} - \int_x^{\frac{\pi}{2}} f(t) (-\csc t \cot t) \, dt \]
This simplifies to:
\[ I_1 = f\left(\frac{\pi}{2}\right) \cdot 1 - f(x) \csc x + \int_x^{\frac{\pi}{2}} f(t) \csc t \cot t \, dt \]
Since \( \csc\left(\frac{\pi}{2}\right) = 1 \) and \( f\left(\frac{\pi}{2}\right) = 3 \):
\[ I_1 = 3 - f(x) \csc x + \int_x^{\frac{\pi}{2}} f(t) \csc t \cot t \, dt \]

### Step 4: Substitute \( I_1 \) into \( g(x) \)
Substituting \( I_1 \) back into \( g(x) \):
\[ g(x) = \left(3 - f(x) \csc x + \int_x^{\frac{\pi}{2}} f(t) \csc t \cot t \, dt\right) - I_2 \]
We can see that the \( I_2 \) term will cancel out with the corresponding term from \( I_1 \).

### Step 5: Finding the Limit as \( x \to 0 \)
Now we need to find:
\[ \lim_{x \to 0} g(x) = \lim_{x \to 0} \left(3 - \frac{f(x)}{\sin x}\right) \]
As \( x \to 0 \), both \( f(x) \) and \( \sin x \) approach 0, leading to a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{f(x)}{\sin x} = \lim_{x \to 0} \frac{f'(x)}{\cos x} \]
Substituting \( x = 0 \):
\[ = \frac{f'(0)}{\cos(0)} = \frac{1}{1} = 1 \]

### Final Calculation
Thus,
\[ \lim_{x \to 0} g(x) = 3 - 1 = 2 \]

### Conclusion
The final answer is:
\[ \lim_{x \to 0} g(x) = 2 \]