Q.Evaluate $\lim_{x \to \frac{1}{2}} \dfrac{4x^2 - 1}{2x - 1}$.

Concept: Limit of a Polynomial (with an indeterminate form that cancels).

The expression $\frac{4x^2 - 1}{2x - 1}$ gives $\frac{0}{0}$ when $x = \frac{1}{2}$, so we must simplify first.

Factor the numerator as a difference of squares:

Cancel the common factor $(2x - 1)$ (valid for $x \neq \frac{1}{2}$):

Now take the limit of the simplified polynomial:

The key idea here is that when a limit gives an indeterminate form like $\frac{0}{0}$, the function often has a removable discontinuity — a hole — at that point. Our job is to simplify the expression so the hole is "filled in" and we can evaluate the limit directly.

For a polynomial divided by a polynomial, if plugging in the limit point gives $\frac{0}{0}$, it means both numerator and denominator share a common factor that becomes zero. Factor both, cancel that factor, and then substitute.

1. Check direct substitution

   Put $x = \frac12$ into the expression:

   Numerator: $4\left(\frac12\right)^2 - 1 = 4 \cdot \frac14 - 1 = 1 - 1 = 0$

   Denominator: $2\left(\frac12\right) - 1 = 1 - 1 = 0$

   So we get $\frac{0}{0}$, an indeterminate form. This tells us the limit exists (if the hole is removable) and we need to simplify.

2. Factor the numerator

   The numerator is $4x^2 - 1$. This is a difference of squares:

3. Cancel the common factor The denominator is $2x - 1$. So the whole expression becomes:

For $x \neq \frac12$, we can cancel $2x - 1$ (since it's not zero), giving:

4. Take the limit Now the limit is easy:

Substitute $x = \frac12$: