If `f (x) = x^(4) + 2x^(3)`, them `lim_(x to 2) (f(x) - f(2))/(x - 2)` is equal to

Options

A. 48
B. 56
C. 64
D. 98

Answer

To solve the limit problem, we need to find:

\[ \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \]

where \( f(x) = x^4 + 2x^3 \).

### Step 1: Calculate \( f(2) \)

First, we need to evaluate \( f(2) \):

\[ f(2) = 2^4 + 2 \cdot 2^3 = 16 + 2 \cdot 8 = 16 + 16 = 32 \]

### Step 2: Set up the limit expression

Now, substituting \( f(2) \) into the limit expression, we have:

\[ \lim_{x \to 2} \frac{f(x) - 32}{x - 2} \]

### Step 3: Substitute \( f(x) \)

Substituting \( f(x) \):

\[ \lim_{x \to 2} \frac{x^4 + 2x^3 - 32}{x - 2} \]

### Step 4: Check for indeterminate form

Now, if we substitute \( x = 2 \) directly into the expression:

\[ f(2) - 32 = 32 - 32 = 0 \]
and
\[ x - 2 = 2 - 2 = 0 \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 5: Apply L'Hôpital's Rule

Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator:

1. Differentiate the numerator:
\[ \frac{d}{dx}(x^4 + 2x^3 - 32) = 4x^3 + 6x^2 \]
2. Differentiate the denominator:
\[ \frac{d}{dx}(x - 2) = 1 \]

### Step 6: Rewrite the limit

Now, we rewrite the limit using the derivatives:

\[ \lim_{x \to 2} \frac{4x^3 + 6x^2}{1} \]

### Step 7: Substitute \( x = 2 \)

Now we can substitute \( x = 2 \):

\[ 4(2^3) + 6(2^2) = 4(8) + 6(4) = 32 + 24 = 56 \]

### Final Answer

Thus, the limit is:

\[ \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} = 56 \]