If $$\lim_{x \to 1}\frac{x^4 - 1}{x - 1} = \lim_{x \to k}\frac{x^3 - k^3}{x^2 - k^2}$$, then k is

We have to equate two limits:

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

We first evaluate the limit on the left. A standard algebraic technique for limits of the type $$\frac{f(x)-f(a)}{x-a}$$ is to factor the numerator. We factor $$x^4-1$$ using the identity $$A^4-B^4=(A-B)(A^3+A^2B+AB^2+B^3).$$ Putting $$A=x$$ and $$B=1$$, we get

$$x^4-1=(x-1)\bigl(x^3+x^2+x+1\bigr).$$

Hence

$$\frac{x^4-1}{x-1}=x^3+x^2+x+1.$$

Now we let $$x \to 1$$:

$$x^3+x^2+x+1 \;\xrightarrow[x\to 1]{}\;1^3+1^2+1+1=1+1+1+1=4.$$

So

$$\lim_{x \to 1}\frac{x^4 - 1}{x - 1}=4.$$

Next we evaluate the limit on the right:

$$\lim_{x \to k}\frac{x^3 - k^3}{x^2 - k^2}.$$

Again we factor numerator and denominator. For the numerator we use $$A^3-B^3=(A-B)(A^2+AB+B^2).$$ With $$A=x$$ and $$B=k$$,

$$x^3-k^3=(x-k)\bigl(x^2+kx+k^2\bigr).$$

The denominator factors by the difference of squares formula $$A^2-B^2=(A-B)(A+B):$$

$$x^2-k^2=(x-k)(x+k).$$

Substituting these factorizations,

$$\frac{x^3-k^3}{x^2-k^2}=\frac{(x-k)(x^2+kx+k^2)}{(x-k)(x+k)}.$$

Because $$x \to k,$$ but $$x \neq k$$ inside the limit, we may cancel the common factor $$x-k$$:

$$\frac{x^3-k^3}{x^2-k^2}=\frac{x^2+kx+k^2}{x+k}.$$

Now we let $$x \to k$$. Replace every occurrence of $$x$$ by $$k$$:

$$\frac{k^2+kk+k^2}{k+k}=\frac{k^2+k^2+k^2}{2k}=\frac{3k^2}{2k}=\frac{3k}{2}.$$

Therefore

$$\lim_{x \to k}\frac{x^3 - k^3}{x^2 - k^2}=\frac{3k}{2}.$$

By the given equality of the two limits, we set

$$4=\frac{3k}{2}.$$

We solve for $$k$$ step by step. Multiply both sides by 2:

$$8=3k.$$

Now divide by 3:

$$k=\frac{8}{3}.$$

Hence, the correct answer is Option 4.