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

Explanation

To solve the limit problem given by

\[ \lim_{x \to 0} \frac{ae^{2x} - b\cos x + c}{x \sin x} = 1, \]

we will follow these steps:

### Step 1: Evaluate the limit at \( x = 0 \)

First, we substitute \( x = 0 \) into the expression:

- \( e^{2x} \) approaches \( e^0 = 1 \)
- \( \cos x \) approaches \( \cos(0) = 1 \)
- \( \sin x \) approaches \( \sin(0) = 0 \)

Thus, the expression becomes:

\[ \frac{a(1) - b(1) + c}{0} = \frac{a - b + c}{0}. \]

This indicates that the limit is in the form \( \frac{0}{0} \), which is indeterminate.

### Step 2: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator separately.

**Numerator:**
- The derivative of \( ae^{2x} \) is \( 2ae^{2x} \).
- The derivative of \( -b\cos x \) is \( b\sin x \).
- The derivative of \( c \) is \( 0 \).

Thus, the derivative of the numerator is:

\[ 2ae^{2x} + b\sin x. \]

**Denominator:**
- The derivative of \( x \sin x \) is \( \sin x + x\cos x \).

### Step 3: Rewrite the limit using derivatives

Now we rewrite the limit:

\[ \lim_{x \to 0} \frac{2ae^{2x} + b\sin x}{\sin x + x\cos x}. \]

### Step 4: Substitute \( x = 0 \) again

Substituting \( x = 0 \) again:

- The numerator becomes \( 2a(1) + b(0) = 2a \).
- The denominator becomes \( 0 + 0(1) = 0 \).

This again gives us a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 5: Differentiate again

**Numerator:**
- The derivative of \( 2ae^{2x} \) is \( 4ae^{2x} \).
- The derivative of \( b\sin x \) is \( b\cos x \).

Thus, the new numerator is:

\[ 4ae^{2x} + b\cos x. \]

**Denominator:**
- The derivative of \( \sin x + x\cos x \) is \( \cos x + (\cos x - x\sin x) = 2\cos x - x\sin x \).

### Step 6: Rewrite the limit again

Now we have:

\[ \lim_{x \to 0} \frac{4ae^{2x} + b\cos x}{2\cos x - x\sin x}. \]

### Step 7: Substitute \( x = 0 \) once more

Substituting \( x = 0 \):

- The numerator becomes \( 4a(1) + b(1) = 4a + b \).
- The denominator becomes \( 2(1) - 0 = 2 \).

Thus, we have:

\[ \frac{4a + b}{2} = 1. \]

### Step 8: Solve for \( a \) and \( b \)

Multiplying both sides by 2 gives:

\[ 4a + b = 2. \]

### Step 9: Use the original limit condition

From the first limit condition, we also have:

\[ a - b + c = 0. \]

### Step 10: Solve the system of equations

Now we have two equations:

1. \( 4a + b = 2 \)
2. \( a - b + c = 0 \)

From equation 1, we can express \( b \) in terms of \( a \):

\[ b = 2 - 4a. \]

Substituting this into equation 2:

\[ a - (2 - 4a) + c = 0 \implies a - 2 + 4a + c = 0 \implies 5a + c = 2. \]

### Step 11: Solve for \( c \)

Now we can express \( c \):

\[ c = 2 - 5a. \]

### Step 12: Find \( a + b + c \)

Now we can find \( a + b + c \):

\[ a + b + c = a + (2 - 4a) + (2 - 5a) = a + 2 - 4a + 2 - 5a = -8a + 4. \]

### Step 13: Find \( a \)

To find \( a \), we can set \( a = 0 \) (from previous deductions). Substituting \( a = 0 \):

\[ b = 2 - 4(0) = 2, \]
\[ c = 2 - 5(0) = 2. \]

### Final Calculation

Thus, we have:

\[ a + b + c = 0 + 2 + 2 = 4. \]

### Conclusion

The value of \( a + b + c \) is

\[ \boxed{4}. \]