`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 problem, we need to evaluate the following limits:

1. \( L_1 = \lim_{x \to 0} \frac{\log(1 + 2x)}{x} \)
2. \( L_2 = \lim_{x \to 2} \frac{x^4 - 2^4}{x - 2} \)

Finally, we will add \( L_1 \) and \( L_2 \).

### Step 1: Evaluate \( L_1 \)

We start with:

\[ L_1 = \lim_{x \to 0} \frac{\log(1 + 2x)}{x} \]

Substituting \( x = 0 \) gives us \( \frac{\log(1 + 0)}{0} = \frac{0}{0} \), which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule.

Using L'Hôpital's Rule:

\[ L_1 = \lim_{x \to 0} \frac{\frac{d}{dx}[\log(1 + 2x)]}{\frac{d}{dx}[x]} \]

Calculating the derivatives:

- The derivative of \( \log(1 + 2x) \) is \( \frac{2}{1 + 2x} \).
- The derivative of \( x \) is \( 1 \).

So we have:

\[ L_1 = \lim_{x \to 0} \frac{2}{1 + 2x} \]

Now substituting \( x = 0 \):

\[ L_1 = \frac{2}{1 + 0} = 2 \]

### Step 2: Evaluate \( L_2 \)

Next, we evaluate:

\[ L_2 = \lim_{x \to 2} \frac{x^4 - 2^4}{x - 2} \]

Substituting \( x = 2 \) gives us \( \frac{0}{0} \), which is again an indeterminate form. We apply L'Hôpital's Rule again.

Using L'Hôpital's Rule:

\[ L_2 = \lim_{x \to 2} \frac{\frac{d}{dx}[x^4 - 2^4]}{\frac{d}{dx}[x - 2]} \]

Calculating the derivatives:

- The derivative of \( x^4 - 2^4 \) is \( 4x^3 \).
- The derivative of \( x - 2 \) is \( 1 \).

Thus:

\[ L_2 = \lim_{x \to 2} \frac{4x^3}{1} \]

Now substituting \( x = 2 \):

\[ L_2 = 4 \cdot 2^3 = 4 \cdot 8 = 32 \]

### Step 3: Add \( L_1 \) and \( L_2 \)

Now we can find the total limit:

\[ L_1 + L_2 = 2 + 32 = 34 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{34} \]