Q.$\lim_{x \to 1} \dfrac{\left(\sqrt{x} - 1\right)(2x - 3)}{2x^2 + x - 3}$ is
(A) $\dfrac{1}{10}$
(B) $\dfrac{-1}{10}$
(C) $1$
(D) None of these

The key idea is that direct substitution gives a $\frac{0}{0}$ form, so we factor and simplify the rational expression.

First, factor the denominator: $2x^2 + x - 3 = (2x + 3)(x - 1)$.

Notice that $x - 1 = (\sqrt{x} - 1)(\sqrt{x} + 1)$. So the numerator becomes $(\sqrt{x} - 1)(2x - 3)$.

Cancel the common factor $(\sqrt{x} - 1)$:

The key idea here is that when a limit gives $\frac{0}{0}$, it means the numerator and denominator share a factor that becomes zero at the limit point. Our job is to find and cancel that factor — then the limit becomes straightforward.

Let’s look at the expression:

The numerator already has $\sqrt{x} - 1$, which is zero at $x=1$. The denominator $2x^2 + x - 3$ also becomes zero at $x=1$ (since $2+1-3=0$). So we have a $\frac{0}{0}$ form — cancellation is possible.

1. Factor the denominator $2x^2 + x - 3$ is a quadratic. We look for two numbers whose product is $2 \times (-3) = -6$ and sum is $1$ (the coefficient of $x$). Those numbers are $3$ and $-2$. So:

2. Rewrite the limit

3. Connect $\sqrt{x} - 1$ to $x-1$ This is the crucial step. Recall the identity:

So $\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$.

Substitute this into the numerator:

4. Cancel the common factor $x-1$ Since $x \neq 1$ in the limit (we only approach 1), $x-1 \neq 0$ and cancellation is valid: …