Q.Evaluate $\displaystyle\lim_{x\to 1}\frac{x^{2} + 2x - 3}{x - 1}$.

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 $x \neq 2$):

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

Step 5 — Take the limit:

$\lim_{x\to2} \frac{x-3}{x(x-1)} = \frac{2-3}{2(2-1)} = \frac{-1}{2} = -\frac{1}{2}.$