`lim_(xto0)(ae^(2x)-bcosx+c)/(xsinx)=1` then `a+b+c` is equal to .__________

Explanation

To solve the limit problem \( \lim_{x \to 0} \frac{ae^{2x} - b\cos x + c}{x \sin x} = 1 \), we will follow these steps:

### Step 1: Rewrite the limit
We can rewrite the limit as:
\[ \lim_{x \to 0} \frac{ae^{2x} - b\cos x + c}{x \sin x} = \lim_{x \to 0} \frac{ae^{2x} - b\cos x + c}{x^2} \cdot \frac{x^2}{\sin x} \]
Since \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we have:
\[ \lim_{x \to 0} \frac{x^2}{\sin x} = 0 \]
Thus, we need to focus on the limit of the numerator.

### Step 2: Expand the functions using Taylor series
Using Taylor series expansions around \( x = 0 \):
- \( e^{2x} \approx 1 + 2x + \frac{(2x)^2}{2} = 1 + 2x + 2x^2 \)
- \( \cos x \approx 1 - \frac{x^2}{2} \)

Substituting these expansions into the limit gives:
\[ ae^{2x} \approx a(1 + 2x + 2x^2) = a + 2ax + 2ax^2 \]
\[ -b\cos x \approx -b(1 - \frac{x^2}{2}) = -b + \frac{bx^2}{2} \]
Thus, the numerator becomes:
\[ ae^{2x} - b\cos x + c \approx (a - b + c) + (2a)x + \left(2a + \frac{b}{2}\right)x^2 \]

### Step 3: Combine terms
Now, we combine the terms:
\[ \text{Numerator} \approx (a - b + c) + (2a)x + \left(2a + \frac{b}{2}\right)x^2 \]

### Step 4: Set the limit equal to 1
For the limit to equal 1, the constant term in the numerator must be 0 (since the denominator approaches 0), and the coefficient of \( x^2 \) must equal 1:
1. \( a - b + c = 0 \)
2. \( 2a + \frac{b}{2} = 1 \)

### Step 5: Solve the equations
From the first equation:
\[ c = b - a \]

Substituting \( c \) into the second equation:
\[ 2a + \frac{b}{2} = 1 \]
Now substituting \( c = b - a \):
\[ 2a + \frac{b}{2} = 1 \]

### Step 6: Solve for \( a \) and \( b \)
From \( 2a + \frac{b}{2} = 1 \):
Multiply through by 2:
\[ 4a + b = 2 \quad \text{(Equation 1)} \]

Now, substituting \( c = b - a \) into the first equation:
\[ a - b + (b - a) = 0 \Rightarrow 0 = 0 \quad \text{(True)} \]

### Step 7: Substitute values
From Equation 1:
\[ b = 2 - 4a \]
Substituting \( b \) into \( c = b - a \):
\[ c = (2 - 4a) - a = 2 - 5a \]

### Step 8: Find \( a + b + c \)
Now, we can find \( a + b + c \):
\[ a + b + c = a + (2 - 4a) + (2 - 5a) = 4 - 8a \]

### Step 9: Set \( a = 0 \) (from the limit condition)
From the limit condition, we find \( a = 0 \):
\[ a + b + c = 4 - 8(0) = 4 \]

Thus, the final answer is:
\[ \boxed{4} \]