`lim_(xrarr0)(2log(1+x)-log(1+2x))/(x^2)` is equal to

Options

A. 0
B. 1
C. -1
D. 2

Explanation

To solve the limit \( \lim_{x \to 0} \frac{2 \log(1+x) - \log(1+2x)}{x^2} \), we will follow these steps:

### Step 1: Rewrite the expression
We can use the property of logarithms that states \( a \log(b) = \log(b^a) \). Thus, we can rewrite \( 2 \log(1+x) \) as:
\[ 2 \log(1+x) = \log((1+x)^2) \]
So, the limit becomes:
\[ \lim_{x \to 0} \frac{\log((1+x)^2) - \log(1+2x)}{x^2} \]

### Step 2: Combine the logarithms
Using the property of logarithms that states \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \), we can combine the logarithms:
\[ \lim_{x \to 0} \frac{\log\left(\frac{(1+x)^2}{1+2x}\right)}{x^2} \]

### Step 3: Apply L'Hôpital's Rule
As \( x \to 0 \), both the numerator and denominator approach 0, creating a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\frac{d}{dx} \left[\log\left(\frac{(1+x)^2}{1+2x}\right)\right]}{\frac{d}{dx}(x^2)} \]

### Step 4: Differentiate the numerator and denominator
1. Differentiate the numerator:
\[ \frac{d}{dx} \left[\log\left(\frac{(1+x)^2}{1+2x}\right)\right] = \frac{1}{\frac{(1+x)^2}{1+2x}} \cdot \frac{d}{dx}\left[\frac{(1+x)^2}{1+2x}\right] \]
Using the quotient rule:
\[ \frac{d}{dx}\left[\frac{(1+x)^2}{1+2x}\right] = \frac{(1+2x)(2(1+x)) - (1+x)^2(2)}{(1+2x)^2} \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(x^2) = 2x \]

### Step 5: Evaluate the limit again
Now substituting back into the limit:
\[ \lim_{x \to 0} \frac{\frac{1}{\frac{(1+x)^2}{1+2x}} \cdot \frac{(1+2x)(2(1+x)) - (1+x)^2(2)}{(1+2x)^2}}{2x} \]

### Step 6: Simplify and evaluate
As \( x \to 0 \):
- The numerator simplifies to \( \frac{2(1)(1)}{1} = 2 \)
- The denominator approaches \( 2(0) = 0 \)

Thus, we can evaluate the limit:
\[ \lim_{x \to 0} \frac{2}{2x} = 1 \]

### Final Result
The limit is:
\[ \lim_{x \to 0} \frac{2 \log(1+x) - \log(1+2x)}{x^2} = 1 \]