`lim_(x to 0) (log (1 + 2x))/(x) + lim_(x to 0) (x^(4) - 2^(4))/(x - 2)` equals

Options

A. 30
B. 32
C. 35
D. 34

Answer

To solve the given limit problem, we will break it down into two parts and evaluate each limit separately.

### Step 1: Evaluate the first limit
We need to evaluate:
\[ \lim_{x \to 0} \frac{\log(1 + 2x)}{x} \]
When we substitute \(x = 0\), we get:
\[ \frac{\log(1 + 2 \cdot 0)}{0} = \frac{\log(1)}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule. According to L'Hôpital's Rule, we differentiate the numerator and the denominator.

**Differentiating the numerator:**
\[ \frac{d}{dx}[\log(1 + 2x)] = \frac{2}{1 + 2x} \]

**Differentiating the denominator:**
\[ \frac{d}{dx}[x] = 1 \]

Now we apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\log(1 + 2x)}{x} = \lim_{x \to 0} \frac{2}{1 + 2x} \]
Substituting \(x = 0\):
\[ \frac{2}{1 + 2 \cdot 0} = \frac{2}{1} = 2 \]

### Step 2: Evaluate the second limit
Next, we evaluate:
\[ \lim_{x \to 2} \frac{x^4 - 2^4}{x - 2} \]
Substituting \(x = 2\), we get:
\[ \frac{2^4 - 2^4}{2 - 2} = \frac{0}{0} \]
This is again an indeterminate form, so we apply L'Hôpital's Rule.

**Differentiating the numerator:**
\[ \frac{d}{dx}[x^4 - 2^4] = 4x^3 \]

**Differentiating the denominator:**
\[ \frac{d}{dx}[x - 2] = 1 \]

Now we apply L'Hôpital's Rule:
\[ \lim_{x \to 2} \frac{x^4 - 2^4}{x - 2} = \lim_{x \to 2} \frac{4x^3}{1} \]
Substituting \(x = 2\):
\[ 4 \cdot (2^3) = 4 \cdot 8 = 32 \]

### Step 3: Combine the results
Now we combine the results of both limits:
\[ \lim_{x \to 0} \frac{\log(1 + 2x)}{x} + \lim_{x \to 2} \frac{x^4 - 2^4}{x - 2} = 2 + 32 = 34 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{34} \]