`lim_(xrarr0) (ax+x cos x)/(b sinx)`

Explanation

To solve the limit \( \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} \]

### Step 2: Factor out \( x \) from the numerator
We can factor \( x \) out of the numerator:
\[ \lim_{x \to 0} \frac{x(a + \cos x)}{b \sin x} \]

### Step 3: Divide numerator and denominator by \( x \)
Next, we divide both the numerator and the denominator by \( x \):
\[ \lim_{x \to 0} \frac{a + \cos x}{b \frac{\sin x}{x}} \]

### Step 4: Use the limit property of \( \frac{\sin x}{x} \)
We know that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Thus, we can substitute this into our limit:
\[ \lim_{x \to 0} \frac{a + \cos x}{b \cdot 1} = \frac{a + \cos x}{b} \]

### Step 5: Evaluate the limit as \( x \to 0 \)
Now we can evaluate the limit by substituting \( x = 0 \):
\[ \frac{a + \cos(0)}{b} = \frac{a + 1}{b} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} = \frac{a + 1}{b} \]