Evaluate the limits, if exist `lim_(x rarr 0)(log_e(1+2x))/(x)`.

Explanation

To evaluate the limit \( \lim_{x \to 0} \frac{\log_e(1 + 2x)}{x} \), we will follow these steps:

### Step 1: Identify the Form
First, we substitute \( x = 0 \) into the expression:
\[ \frac{\log_e(1 + 2 \cdot 0)}{0} = \frac{\log_e(1)}{0} = \frac{0}{0} \]
This is an indeterminate form (0/0), so we can apply L'Hôpital's Rule or use the Taylor series expansion.

### Step 2: Use the Taylor Series Expansion
The Taylor series expansion for \( \log_e(1 + x) \) around \( x = 0 \) is:
\[ \log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
For our case, we substitute \( x \) with \( 2x \):
\[ \log_e(1 + 2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \ldots \]
This simplifies to:
\[ \log_e(1 + 2x) = 2x - 2x^2 + \frac{8x^3}{3} - \ldots \]

### Step 3: Substitute into the Limit
Now we substitute this expansion back into our limit:
\[ \lim_{x \to 0} \frac{\log_e(1 + 2x)}{x} = \lim_{x \to 0} \frac{2x - 2x^2 + \frac{8x^3}{3} - \ldots}{x} \]
This simplifies to:
\[ \lim_{x \to 0} \left( 2 - 2x + \frac{8x^2}{3} - \ldots \right) \]

### Step 4: Evaluate the Limit
As \( x \to 0 \), all terms containing \( x \) vanish:
\[ \lim_{x \to 0} (2 - 2x + \frac{8x^2}{3} - \ldots) = 2 \]

### Conclusion
Thus, we find that:
\[ \lim_{x \to 0} \frac{\log_e(1 + 2x)}{x} = 2 \]