Applying Largrange's mean value theorem for a suitable function `f (x)` in `[0,h],` we have
`f (h) =f (0) + hf'(thetah) , 0 lt theta lt 1.`
Then for `f (x) = cos x,` the vlaue of `lim _( h to 0 ^(+)) theta ` is

Options

A. 1
B. 0
C. `1//2`
D. `1//3`

Answer

To solve the problem using Lagrange's Mean Value Theorem, we will follow these steps:

### Step 1: Define the function
We are given the function \( f(x) = \cos x \).

### Step 2: Apply Lagrange's Mean Value Theorem
According to Lagrange's Mean Value Theorem, for a continuous function on the interval \([a, b]\) and differentiable on \((a, b)\), there exists a point \( \theta \) in \((0, 1)\) such that:
\[ f(b) = f(a) + (b - a) f'(\theta) \]
In our case, we set \( a = 0 \) and \( b = h \):
\[ f(h) = f(0) + h f'(\theta h) \]

### Step 3: Calculate \( f(0) \) and \( f(h) \)
We know:
\[ f(0) = \cos(0) = 1 \]
\[ f(h) = \cos(h) \]
Substituting these into the equation gives:
\[ \cos(h) = 1 + h f'(\theta h) \]

### Step 4: Find the derivative \( f'(x) \)
The derivative of \( f(x) = \cos x \) is:
\[ f'(x) = -\sin x \]
Thus,
\[ f'(\theta h) = -\sin(\theta h) \]

### Step 5: Substitute the derivative back into the equation
Now substituting \( f'(\theta h) \) into our equation:
\[ \cos(h) = 1 - h \sin(\theta h) \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \cos(h) - 1 = -h \sin(\theta h) \]
This can be rewritten as:
\[ \frac{\cos(h) - 1}{h} = -\sin(\theta h) \]

### Step 7: Evaluate the limit as \( h \to 0^+ \)
We need to evaluate:
\[ \lim_{h \to 0^+} \theta \]
To find \( \theta \), we can express it in terms of \( h \):
\[ \sin(\theta h) = \frac{\cos(h) - 1}{-h} \]
As \( h \to 0 \), we know from Taylor series expansion:
\[ \cos(h) \approx 1 - \frac{h^2}{2} \]
Thus:
\[ \cos(h) - 1 \approx -\frac{h^2}{2} \]
Substituting this back:
\[ \frac{-\frac{h^2}{2}}{h} = -\frac{h}{2} \]
So:
\[ \sin(\theta h) \approx -\frac{h}{2} \]

### Step 8: Use the small angle approximation
For small angles, we have \( \sin x \approx x \). Therefore:
\[ \theta h \approx -\frac{h}{2} \]
This implies:
\[ \theta \approx -\frac{1}{2} \]

### Step 9: Evaluate the limit
Taking the limit as \( h \to 0^+ \):
\[ \lim_{h \to 0^+} \theta = -\frac{1}{2} \]

### Final Answer
Thus, the value of \( \lim_{h \to 0^+} \theta \) is:
\[ \boxed{-\frac{1}{2}} \]