Q.If $\lim_{x \to 1} \dfrac{x^4 - 1}{x - 1} = \lim_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2}$, then find the value of $k$.

Concept: Limit of polynomial quotients using factorization and direct substitution.

Both limits have indeterminate form $\frac{0}{0}$, so we factor and simplify.

Left side:

Right side: …

When you see a limit of the form $\frac{f(x) - f(a)}{x - a}$ as $x \to a$, you're looking at the definition of the derivative $f'(a)$. The left-hand side is exactly this pattern, and the right-hand side can be massaged into the same form. Once we recognize both as derivatives, the problem becomes an equation between two numbers.

Why this approach works

The limit $\lim_{x \to 1} \frac{x^4 - 1}{x - 1}$ is the derivative of $x^4$ at $x = 1$. For the right-hand side, we need to rewrite $\frac{x^3 - k^3}{x^2 - k^2}$ in a form that reveals its derivative structure. Both limits exist (they're not indeterminate disasters), so we can evaluate them and set them equal.

Solution

1. Evaluate the left-hand limit.

   We have $\lim_{x \to 1} \frac{x^4 - 1}{x - 1}$. This is the derivative of $f(x) = x^4$ at $x = 1$:

2. Rewrite the right-hand limit.

   The expression $\frac{x^3 - k^3}{x^2 - k^2}$ as $x \to k$ is a $\frac{0}{0}$ form. Factor both numerator and denominator:

So for $x \neq k$:

3. Evaluate the right-hand limit. …