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

Answer

To solve the problem, we need to find the value of \( k \) such that

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}. \]

### Step 1: Calculate the left-hand limit

First, we evaluate the limit as \( x \) approaches 1 for the expression \( \frac{x^4 - 1}{x - 1} \).

The expression \( x^4 - 1 \) can be factored as:

\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1). \]

Thus, we can rewrite the limit:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1}. \]

Since \( x - 1 \) cancels out, we have:

\[ \lim_{x \to 1} (x + 1)(x^2 + 1). \]

Now substituting \( x = 1 \):

\[ (1 + 1)(1^2 + 1) = 2 \cdot 2 = 4. \]

### Step 2: Set up the right-hand limit

Next, we evaluate the limit as \( x \) approaches \( k \) for the expression \( \frac{x^3 - k^3}{x^2 - k^2} \).

Using the difference of cubes and the difference of squares, we can rewrite the limit:

\[ x^3 - k^3 = (x - k)(x^2 + xk + k^2), \]
\[ x^2 - k^2 = (x - k)(x + k). \]

Thus, we have:

\[ \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{(x - k)(x^2 + xk + k^2)}{(x - k)(x + k)}. \]

Again, \( x - k \) cancels out, leading to:

\[ \lim_{x \to k} \frac{x^2 + xk + k^2}{x + k}. \]

### Step 3: Substitute \( x = k \)

Now substituting \( x = k \):

\[ \frac{k^2 + k^2 + k^2}{k + k} = \frac{3k^2}{2k} = \frac{3k}{2}. \]

### Step 4: Set the limits equal

Now we equate the two limits:

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

### Step 5: Solve for \( k \)

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

\[ 8 = 3k. \]

Now, divide by 3:

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

### Final Answer

Thus, the value of \( k \) is

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