Evaluate: `lim_(x rarr 0) (sin (log (1 + 2x)))/(x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sin(\log(1 + 2x))}{x}, \]

we can use some known limits and properties of logarithms and trigonometric functions.

### Step 1: Rewrite the limit

We start with the limit:

\[ \lim_{x \to 0} \frac{\sin(\log(1 + 2x))}{x}. \]

### Step 2: Use the known limit for sine

We know that

\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1. \]

To use this, we need to express our limit in a form that allows us to apply this limit. We can rewrite the limit as follows:

\[ \lim_{x \to 0} \frac{\sin(\log(1 + 2x))}{\log(1 + 2x)} \cdot \frac{\log(1 + 2x)}{x}. \]

### Step 3: Analyze the first part

Now, we can analyze the first part:

\[ \lim_{x \to 0} \frac{\sin(\log(1 + 2x))}{\log(1 + 2x)}. \]

As \(x \to 0\), \(\log(1 + 2x) \to \log(1) = 0\), so we can apply the known limit:

\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1. \]

Thus,

\[ \lim_{x \to 0} \frac{\sin(\log(1 + 2x))}{\log(1 + 2x)} = 1. \]

### Step 4: Analyze the second part

Next, we need to evaluate the second part:

\[ \lim_{x \to 0} \frac{\log(1 + 2x)}{x}. \]

Using the known limit

\[ \lim_{x \to 0} \frac{\log(1 + x)}{x} = 1, \]

we can adapt this for \(\log(1 + 2x)\):

\[ \lim_{x \to 0} \frac{\log(1 + 2x)}{x} = \lim_{x \to 0} \frac{\log(1 + 2x)}{2x} \cdot 2 = 2. \]

### Step 5: Combine the results

Now we can combine our results:

\[ \lim_{x \to 0} \frac{\sin(\log(1 + 2x))}{x} = \lim_{x \to 0} \left( \frac{\sin(\log(1 + 2x))}{\log(1 + 2x)} \cdot \frac{\log(1 + 2x)}{x} \right) = 1 \cdot 2 = 2. \]

### Final Answer

Thus, the limit is

\[ \boxed{2}. \]