Q.Prove that the function defined by $f(x) = \tan x$ is a continuous function.

The key idea is that continuity at a point requires the limit of the function as $x$ approaches that point to equal the function's value there. For $f(x) = \tan x$, we use the fact that it is the quotient of two continuous functions.

1. Write $\tan x = \frac{\sin x}{\cos x}$. Both $\sin x$ and $\cos x$ are continuous for all real $x$ (a standard result from their definitions as limits of series or from geometric reasoning).
2. The quotient of two continuous functions is continuous wherever the denominator is non-zero. Here, $\cos x = 0$ at $x = \frac{\pi}{2} + n\pi$ for any integer $n$. …

The Core Idea: Continuity of a Quotient

When we say a function is "continuous," we mean it is continuous at every point in its domain. For $f(x) = \tan x$, the domain is all real numbers except where $\cos x = 0$ — that is, all $x \neq \frac{\pi}{2} + n\pi$ for any integer $n$.

The key insight is that $\tan x$ is built from simpler functions:

Both $\sin x$ and $\cos x$ are continuous everywhere on $\mathbb{R}$. This is a standard result from calculus — you can prove it using the limit definition, but for our purposes we take it as given.

Now, there's a powerful theorem about continuity: if two functions are continuous at a point, their quotient is also continuous at that point, provided the denominator is non-zero there. That's exactly the situation here.

So the proof reduces to checking two things:

1. Are $\sin x$ and $\cos x$ continuous? Yes, everywhere.
2. Where is $\cos x \neq 0$? Everywhere except $x = \frac{\pi}{2} + n\pi$.

At those exceptional points, $\tan x$ isn't even defined, so the question of continuity doesn't arise. A function can only be continuous at points in its domain.

Step-by-Step Proof

1. Recall the definition of continuity at a point.

A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. This requires three things: $f(a)$ exists, the limit exists, and they are equal.

2. Express $\tan x$ as a quotient.

We write $f(x) = \frac{\sin x}{\cos x}$. Let $g(x) = \sin x$ and $h(x) = \cos x$.

3. Establish the continuity of $g$ and $h$.

Both $\sin x$ and $\cos x$ are continuous on $\mathbb{R}$. This is a standard result from the epsilon-delta definition of limits, but you can also see it geometrically: as $x$ changes by a small amount, $\sin x$ and $\cos x$ change by a small amount. No jumps, no breaks.

4. Apply the quotient rule for continuity.

Take any point $a$ in the domain of $\tan x$, meaning $\cos a \neq 0$. Since $g$ and $h$ are continuous at $a$ and $h(a) \neq 0$, the quotient $\frac{g}{h}$ is continuous at $a$.

5. Conclude for the entire domain. …

Method: Proving Continuity of a Trigonometric Quotient (e.g., $\tan x = \sin x/\cos x$) on Its Domain

This method applies whenever a function is expressed as the ratio of two standard functions whose continuity is already known, and the domain is restricted by where the denominator vanishes.

Steps

Step 1: Rewrite the function as an explicit quotient of two functions you already know are continuous.

Both $\sin x$ and $\cos x$ are continuous on all of $\mathbb{R}$ — a fact you can cite once it has been established (e.g. by the inequality method above).

Step 2: Determine exactly where the denominator is zero.

Solve $\cos x = 0$; here this happens at $x = \frac{\pi}{2}+n\pi$ for integer $n$. These points are excluded from the domain of $\tan x$ from the outset.

Step 3: Apply the quotient rule for continuity at every remaining point.

$\text{If } g,\ h \text{ continuous at } a, \ h(a)\neq 0 \Rightarrow \frac{g}{h} \text{ continuous at } a.$ …

Common Mistakes

Mistake 1: Describing $\tan x$ as "discontinuous" at $x=\frac{\pi}{2}$.

Why it's wrong: discontinuity, strictly, applies to a point that is in the domain but fails the continuity test — since $\tan\frac{\pi}{2}$ is not defined at all (division by zero), that point isn't part of the domain to begin with, so calling it a discontinuity is imprecise. Correct approach: say the function is "not defined" at $\frac{\pi}{2}+n\pi$, and reserve "continuous on its domain" as the actual claim being proved.

Mistake 2: Forgetting to solve $\cos x = 0$ for the general solution.

Why it's wrong: only listing $x=\frac{\pi}{2}$ and missing the full family $\frac{\pi}{2}+n\pi$ (for all integers $n$) understates the excluded set and makes the domain description incomplete. Correct approach: always give the general solution of the denominator's zero equation, using the periodicity of $\cos x$. …