Q.Evaluate: $\lim\limits_{x \to 2} \left[ \dfrac{1}{x - 2} - \dfrac{2(2x - 3)}{x^3 - 3x^2 + 2x} \right]$

The limit equals $-\dfrac{1}{2}$.

As $x \to 2$ each fraction tends to $\infty$, so this is the indeterminate form $\infty - \infty$; combine them into a single fraction first.

Step 1 — Factorise the denominator of the second term.

$x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x-1)(x-2).$

So the expression is

$\frac{1}{x-2} - \frac{2(2x-3)}{x(x-1)(x-2)}.$

Step 2 — Common denominator $x(x-1)(x-2)$:

$= \frac{x(x-1) - 2(2x-3)}{x(x-1)(x-2)}.$

Step 3 — Simplify the numerator:

$x(x-1) - 2(2x-3) = x^2 - x - 4x + 6 = x^2 - 5x + 6 = (x-2)(x-3).$

Step 4 — Cancel $(x-2)$ (valid for $x \neq 2$): …