Evaluate, `lim_(x to 1) (x^(4)-1)/(x-1)=lim_(x to k) (x^(3)-k^(3))/(x^(2)-k^(2))` , then find the value of k.

Options

A. `5/3`
B. `1/3`
C. `2/3`
D. `8/3`

Explanation

To solve the problem, we need to evaluate the limits on both sides of the equation and find the value of \( k \).

### Step 1: Evaluate the limit on the left side

We start with the limit:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \]

This limit is in the form \( \frac{0}{0} \) when we substitute \( x = 1 \). We can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the denominator.

**Derivative of the numerator**:
\[ \frac{d}{dx}(x^4 - 1) = 4x^3 \]

**Derivative of the denominator**:
\[ \frac{d}{dx}(x - 1) = 1 \]

Now, applying L'Hôpital's Rule:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to 1} \frac{4x^3}{1} = 4(1)^3 = 4 \]

### Step 2: Evaluate the limit on the right side

Next, we evaluate the limit:

\[ \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} \]

This limit is also in the form \( \frac{0}{0} \) when we substitute \( x = k \). We can again apply L'Hôpital's Rule.

**Derivative of the numerator**:
\[ \frac{d}{dx}(x^3 - k^3) = 3x^2 \]

**Derivative of the denominator**:
\[ \frac{d}{dx}(x^2 - k^2) = 2x \]

Now, applying L'Hôpital's Rule:

\[ \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{3x^2}{2x} = \lim_{x \to k} \frac{3x}{2} = \frac{3k}{2} \]

### Step 3: Set the limits equal to each other

Now we set the two limits equal to each other:

\[ 4 = \frac{3k}{2} \]

### Step 4: Solve for \( k \)

To solve for \( k \), we multiply both sides by 2:

\[ 8 = 3k \]

Now, divide both sides by 3:

\[ k = \frac{8}{3} \]

### Final Answer

The value of \( k \) is:

\[ \boxed{\frac{8}{3}} \]