Q.Discuss the continuity of the function $f$ defined by $f(x) = \frac{1}{x}$, $x \neq 0$.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. Here, $f$ is not defined at $x = 0$, so the condition fails.

1. The function $f(x) = \frac{1}{x}$ is defined only for $x \neq 0$.
2. For any $a \neq 0$, $\lim_{x \to a} \frac{1}{x} = \frac{1}{a} = f(a)$, so $f$ is continuous at every $a \neq 0$. …

The Core Idea: Continuity at a Point

Before we check anything, we need the precise definition. A function $f$ is continuous at a point $x = a$ if three conditions hold:

1. $f(a)$ is defined (the point belongs to the domain).
2. $\lim_{x \to a} f(x)$ exists (the two-sided limit is finite).
3. $\lim_{x \to a} f(x) = f(a)$ (the limit equals the function value).

If any one of these fails, the function is discontinuous at that $a$.

The function $f(x) = 1/x$ is one of the first examples students meet. Its graph is a hyperbola with two separate branches — one in the first quadrant, one in the third. The natural question is: where is it continuous?

Step-by-Step Analysis

1. Identify the domain.

The function $f(x) = 1/x$ is defined for all real $x$ except $0$. So the domain is $\mathbb{R} \setminus \{0\}$. Every point $a \neq 0$ is in the domain.

2. Check the limit at a general point $a \neq 0$.

For any $a \neq 0$, we know from the standard limit laws that

Why? Because $g(x) = x$ is continuous everywhere, and the reciprocal function $h(t) = 1/t$ is continuous at every $t \neq 0$. The composition of continuous functions is continuous, but more directly: for $x$ close to $a$, $1/x$ gets arbitrarily close to $1/a$. No drama — the limit exists and is finite.

3. Compare the limit to the function value.

At $x = a$, $f(a) = 1/a$. So

All three conditions hold: $f(a)$ is defined, the limit exists, and they match.

4. What about $x = 0$? …

Method: Discussing Continuity of a Function With a Restricted Domain

This method handles functions (typically rational expressions) that are undefined at certain points, where the task is to describe continuity correctly across the function's actual domain rather than incorrectly declaring a "discontinuity" at a point outside it.

Steps

Step 1: Determine the domain precisely

Find every value of $x$ for which the function's formula is undefined (commonly where a denominator becomes zero) and exclude those from the domain.

Step 2: Check continuity at an arbitrary point inside the domain

For any point $a$ in the domain, use the algebra-of-continuous-functions results (the identity and constant functions are continuous, so a quotient of polynomials is continuous wherever the denominator is non-zero) to conclude

Step 3: Examine the behaviour at an excluded point separately …

Common Mistakes

Mistake 1: Saying "$f(x)=1/x$ is discontinuous at $x=0$"

Why it's wrong: continuity (and discontinuity) is only defined at points that belong to the function's domain; since $x=0$ is excluded from the domain of $1/x$ from the start, it is not a point of discontinuity in the strict definitional sense — it is simply outside the domain. Correct approach: state that the function is continuous throughout its domain ($x \neq 0$) and separately describe $x=0$ as excluded from the domain (with asymptotic behaviour), rather than calling it a discontinuity of the function.

Mistake 2: Trying to force a value at $x=0$ to "fix" the function …