Q.(c) Test the continuity of the following function at $x = 1$: $f(x) = \begin{cases} \dfrac{x^2 - 4x + 3}{x - 1}, & x \neq 1 \\ -2, & x = 1 \end{cases}$

Three conditions: $f(a)$ exists, the limit exists, and the two are equal.

A function $f(x)$ is said to be continuous at $x=a$ if all three of the following conditions are satisfied:

1. $f(a)$ is defined — the function has a definite, finite value at $x=a$.
2. $\lim\limits_{x\to a}f(x)$ exists — the left-hand limit and the right-hand limit are equal and finite:

$\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)$

3. The limit equals the function value:

$\lim_{x\to a}f(x)=f(a)$

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$f(x)$ is continuous at $x = 1$ since $\lim_{x\to1}f(x) = f(1) = -2$.

A function is continuous at $x = a$ if (i) $f(a)$ exists, (ii) $\lim_{x\to a}f(x)$ exists, and (iii) the two are equal.

Functional value: given $f(1) = -2$.

Limit: for $x \neq 1$, factorise the numerator ($x^2 - 4x + 3 = (x-1)(x-3)$):

$f(x) = \frac{(x-1)(x-3)}{x-1} = x - 3 \quad (x \neq 1).$

Hence

$\lim_{x\to1} f(x) = \lim_{x\to1}(x - 3) = 1 - 3 = -2.$

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