If `lim_(xto0) (1-cosx)/(e^(ax)-bx-1)=2` then find the values of a and b.

Explanation

To solve the limit problem, we need to find the values of \( a \) and \( b \) such that:

\[ \lim_{x \to 0} \frac{1 - \cos x}{e^{ax} - bx - 1} = 2 \]

### Step 1: Identify the form of the limit
First, we evaluate the limit as \( x \) approaches 0.

- The numerator \( 1 - \cos(0) = 1 - 1 = 0 \).
- The denominator \( e^{a \cdot 0} - b \cdot 0 - 1 = 1 - 0 - 1 = 0 \).

Since both the numerator and denominator approach 0, we have a \( \frac{0}{0} \) indeterminate form, which allows us to apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

- The derivative of the numerator \( 1 - \cos x \) is \( \sin x \).
- The derivative of the denominator \( e^{ax} - bx - 1 \) is \( ae^{ax} - b \).

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sin x}{ae^{ax} - b} \]

### Step 3: Evaluate the new limit
Now we substitute \( x = 0 \):

- The numerator becomes \( \sin(0) = 0 \).
- The denominator becomes \( ae^{a \cdot 0} - b = a \cdot 1 - b = a - b \).

This gives us another \( \frac{0}{0} \) form if \( a = b \). Therefore, we set:

\[ a - b = 0 \quad \Rightarrow \quad a = b \quad \text{(Equation 1)} \]

### Step 4: Differentiate again
Since we have \( a = b \), we differentiate again. The new limit is:

\[ \lim_{x \to 0} \frac{\cos x}{a \cdot ae^{ax}} = \frac{\cos(0)}{a^2} = \frac{1}{a^2} \]

### Step 5: Set the limit equal to 2
We know from the problem statement that this limit equals 2:

\[ \frac{1}{a^2} = 2 \]

### Step 6: Solve for \( a \)
From the equation \( \frac{1}{a^2} = 2 \), we can solve for \( a^2 \):

\[ a^2 = \frac{1}{2} \]

Taking the square root gives:

\[ a = \pm \frac{1}{\sqrt{2}} \]

### Step 7: Find \( b \)
Since we established that \( a = b \), we have:

\[ b = \pm \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:

\[ a = b = \frac{1}{\sqrt{2}} \quad \text{or} \quad a = b = -\frac{1}{\sqrt{2}} \]