`"lt"_(x to 1) (x^(4)-1)/(x-1) = "lt"_(x to k) (x^(3) -k^(3))/(x^(2) -k^(2))` then k=

Options

A. `2/3`
B. `4/3`
C. `8/3`
D. none

Answer

To solve the limit equation \( \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} \), we will follow these steps:

### Step 1: Evaluate the left-hand limit
We start with the left-hand side:
\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \]
We notice that both the numerator and denominator approach 0 as \( x \to 1 \), resulting in the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and denominator:
- The derivative of the numerator \( x^4 - 1 \) is \( 4x^3 \).
- The derivative of the denominator \( x - 1 \) is \( 1 \).

Now we can rewrite the limit:
\[ \lim_{x \to 1} \frac{4x^3}{1} \]

### Step 3: Substitute \( x = 1 \)
Now we substitute \( x = 1 \):
\[ 4(1)^3 = 4 \]
Thus, the left-hand limit evaluates to 4.

### Step 4: Evaluate the right-hand limit
Next, we evaluate the right-hand side:
\[ \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} \]
Again, we see that substituting \( x = k \) results in the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

### Step 5: Apply L'Hôpital's Rule to the right-hand limit
Differentiating the numerator and denominator:
- The derivative of the numerator \( x^3 - k^3 \) is \( 3x^2 \).
- The derivative of the denominator \( x^2 - k^2 \) is \( 2x \).

Now we rewrite the limit:
\[ \lim_{x \to k} \frac{3x^2}{2x} \]

### Step 6: Simplify and substitute \( x = k \)
We can simplify this to:
\[ \lim_{x \to k} \frac{3x}{2} \]
Substituting \( x = k \):
\[ \frac{3k}{2} \]

### Step 7: Set the two limits equal
Now we set the two limits equal to each other:
\[ 4 = \frac{3k}{2} \]

### Step 8: 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:
\[ \boxed{\frac{8}{3}} \]