Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} \dfrac{2x^2 - 3x - 2}{x - 2}, & x \ne 2 \\ 5, & x = 2 \end{cases}$ at $x = 2$.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: Simplify the expression for $x \ne 2$.

Factor the numerator:

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

Thus for $x \ne 2$,

$f(x) = \dfrac{(2x+1)(x-2)}{x-2} = 2x + 1$.

Step 2: Find the limit as $x \to 2$.

$\displaystyle \lim_{x \to 2} f(x) = \lim_{x \to 2} (2x + 1) = 2(2) + 1 = 5$. …

The Core Idea: Continuity at a Point

A function is continuous at a point $x = a$ if three things happen:

1. The function is defined at $a$ (i.e., $f(a)$ exists).
2. The limit of $f(x)$ as $x$ approaches $a$ exists.
3. That limit equals the function value: $\lim_{x\to a} f(x) = f(a)$.

Here, the function is given in piecewise form. For $x \neq 2$, it's a rational expression that looks like it might blow up (division by zero). But the numerator might also vanish at $x=2$, creating a removable discontinuity — a "hole" that can be filled. The piecewise definition gives $f(2)=5$, which is exactly the value that would fill that hole if the limit matches.

So the entire question boils down to: Does $\lim_{x\to 2} \frac{2x^2 - 3x - 2}{x-2}$ equal 5?

Step-by-Step Solution

1. Check if the function is defined at $x=2$

The piecewise definition directly gives $f(2) = 5$. So condition 1 is satisfied.

2. Examine the limit as $x \to 2$

For $x \neq 2$, we have:

If we naively substitute $x=2$, we get $\frac{0}{0}$, an indeterminate form. This suggests the numerator and denominator share a common factor.

3. Factor the numerator

We need to factor $2x^2 - 3x - 2$. Look for two numbers whose product is $2 \times (-2) = -4$ and sum is $-3$. Those numbers are $-4$ and $+1$.

Rewrite the middle term:

Group:

Factor out $(x-2)$:

4. Cancel the common factor

For $x \neq 2$, we can cancel:

$f(x) = \frac{(x-2)(2x+1)}{x-2} = 2x + 1$ …

Method: Removing a 0/0 Rational-Function Indeterminacy by Factoring

This method applies to any continuity check where the "$x \ne a$" branch is a rational function that gives $\frac{0}{0}$ on direct substitution — a sign that the point in question is a removable discontinuity of the underlying expression.

Steps

Step 1: Confirm the indeterminate form.

Substitute $x = a$ into the given rational expression. If both numerator and denominator vanish, the factor $(x-a)$ divides both — this justifies factoring rather than concluding the limit doesn't exist.

Step 2: Factor the numerator and denominator to expose the common factor.

For a quadratic numerator, use the standard "split the middle term" technique: find two numbers whose product matches (leading coefficient) $\times$ (constant term) and whose sum matches the middle coefficient, then group and factor.

Step 3: Cancel the common factor — valid because $x \ne a$ throughout the limiting process. …

Common Mistakes

Mistake 1: Concluding the function is discontinuous just because direct substitution gives $\frac{0}{0}$.

Why it's wrong: $\frac{0}{0}$ signals an indeterminate form that needs simplification, not automatic non-existence of the limit — many such expressions do have a well-defined limit once the common factor is removed. Correct approach: factor the numerator and cancel the shared factor with the denominator before evaluating the limit.

Mistake 2: Sign or factoring errors while splitting the middle term. …