To solve the limit problem
\[
\lim_{t \to 0} \int_{0}^{2\pi} \frac{|\sin(x+t) - \sin x|}{|t|} \, dx,
\]
we can follow these steps:
### Step 1: Rewrite the expression using the sine difference formula.
Using the sine difference identity, we have:
\[
\sin(x+t) - \sin x = 2 \cos\left(\frac{(x+t) + x}{2}\right) \sin\left(\frac{(x+t) - x}{2}\right) = 2 \cos\left(x + \frac{t}{2}\right) \sin\left(\frac{t}{2}\right).
\]
Thus, we can rewrite the integral as:
\[
\int_{0}^{2\pi} \frac{|\sin(x+t) - \sin x|}{|t|} \, dx = \int_{0}^{2\pi} \frac{2 |\cos(x + \frac{t}{2})| \cdot |\sin(\frac{t}{2})|}{|t|} \, dx.
\]
### Step 2: Simplify the expression inside the limit.
Now, we can simplify the expression further:
\[
\lim_{t \to 0} \int_{0}^{2\pi} \frac{2 |\cos(x + \frac{t}{2})| \cdot |\sin(\frac{t}{2})|}{|t|} \, dx.
\]
As \( t \to 0 \), we know that \( |\sin(\frac{t}{2})| \approx \frac{t}{2} \) (using the small angle approximation). Therefore, we can substitute this into our integral:
\[
\lim_{t \to 0} \int_{0}^{2\pi} \frac{2 |\cos(x + \frac{t}{2})| \cdot \frac{t}{2}}{|t|} \, dx = \lim_{t \to 0} \int_{0}^{2\pi} |\cos(x + \frac{t}{2})| \, dx.
\]
### Step 3: Evaluate the limit.
As \( t \to 0 \), \( \cos(x + \frac{t}{2}) \) approaches \( \cos x \). Thus, we can evaluate the integral:
\[
\int_{0}^{2\pi} |\cos x| \, dx.
\]
### Step 4: Calculate the integral of \( |\cos x| \).
The function \( |\cos x| \) is symmetric and periodic. We can compute it over one period:
\[
\int_{0}^{2\pi} |\cos x| \, dx = 2 \int_{0}^{\pi} \cos x \, dx = 2 \left[ \sin x \right]_{0}^{\pi} = 2(0 - 0) + 2(1 - 0) = 2.
\]
### Step 5: Final result.
Thus, the limit evaluates to:
\[
\lim_{t \to 0} \int_{0}^{2\pi} \frac{|\sin(x+t) - \sin x|}{|t|} \, dx = 4.
\]
### Final Answer:
\[
\boxed{4}.
\]
