`lim_(h to 0)(ln(1+ 2h)-2ln(1+h))/h^(2)` = …………..

Answer

To solve the limit

\[ \lim_{h \to 0} \frac{\ln(1 + 2h) - 2\ln(1 + h)}{h^2}, \]

we first evaluate the expression as \( h \) approaches 0.

### Step 1: Identify the form of the limit
Substituting \( h = 0 \) into the expression gives:

\[ \ln(1 + 2 \cdot 0) - 2\ln(1 + 0) = \ln(1) - 2\ln(1) = 0 - 0 = 0. \]

Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that for limits of the form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator.

#### Derivative of the numerator:
The numerator is \( \ln(1 + 2h) - 2\ln(1 + h) \).

Using the chain rule:
- The derivative of \( \ln(1 + 2h) \) is \( \frac{2}{1 + 2h} \).
- The derivative of \( 2\ln(1 + h) \) is \( \frac{2}{1 + h} \).

Thus, the derivative of the numerator is:

\[ \frac{2}{1 + 2h} - \frac{2}{1 + h}. \]

#### Derivative of the denominator:
The denominator is \( h^2 \), and its derivative is \( 2h \).

### Step 3: Rewrite the limit
Now we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{h \to 0} \frac{\frac{2}{1 + 2h} - \frac{2}{1 + h}}{2h}. \]

### Step 4: Simplify the numerator
We need to simplify the numerator:

\[ \frac{2}{1 + 2h} - \frac{2}{1 + h} = 2\left(\frac{1}{1 + 2h} - \frac{1}{1 + h}\right). \]

Finding a common denominator:

\[ = 2 \cdot \frac{(1 + h) - (1 + 2h)}{(1 + 2h)(1 + h)} = 2 \cdot \frac{1 + h - 1 - 2h}{(1 + 2h)(1 + h)} = 2 \cdot \frac{-h}{(1 + 2h)(1 + h)}. \]

### Step 5: Substitute back into the limit
Now substituting this back into the limit, we have:

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

### Step 6: Cancel \( h \) and evaluate the limit
We can cancel \( h \) (as long as \( h \neq 0 \)):

\[ = \lim_{h \to 0} \frac{-1}{(1 + 2h)(1 + h)}. \]

Now substituting \( h = 0 \):

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

### Final Answer
Thus, the limit is

\[ \boxed{-1}. \]