Evaluate `lim_(x to 0) ((1+x)^(6)-1)/((1+x)^(2)-1)`.

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1}, \]

we will follow these steps:

### Step 1: Substitute the limit directly
First, let's check what happens when we substitute \( x = 0 \):

\[ \frac{(1+0)^6 - 1}{(1+0)^2 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}. \]

Since we have an indeterminate form \( \frac{0}{0} \), we need to manipulate the expression further.

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

- **Numerator**: \( f(x) = (1+x)^6 - 1 \)
- **Denominator**: \( g(x) = (1+x)^2 - 1 \)

Now, we find the derivatives:

1. Derivative of the numerator:
\[ f'(x) = 6(1+x)^5 \]

2. Derivative of the denominator:
\[ g'(x) = 2(1+x) \]

### Step 3: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{6(1+x)^5}{2(1+x)}. \]

### Step 4: Simplify the expression
We can simplify this limit:

\[ = \lim_{x \to 0} \frac{6(1+x)^5}{2(1+x)} = \lim_{x \to 0} \frac{6(1+x)^4}{2} = \lim_{x \to 0} 3(1+x)^4. \]

### Step 5: Evaluate the limit
Now substituting \( x = 0 \):

\[ = 3(1+0)^4 = 3 \cdot 1 = 3. \]

Thus, the final answer is:

\[ \lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1} = 3. \]