Q.Discuss the continuity of sine function.

Concept: Continuity At A Point — check $\displaystyle\lim_{x\to c}\sin x=\sin c$ for an arbitrary $c$.

Put $x=c+h$, so $h\to0$ as $x\to c$. Then

Using $\displaystyle\lim_{h\to0}\sin h=0$ and $\displaystyle\lim_{h\to0}\cos h=1$:

Since $c$ was arbitrary, $\sin x$ is continuous at every real number.

To discuss continuity of $f(x)=\sin x$, take an arbitrary real number $c$ and check whether $\displaystyle\lim_{x\to c}f(x)=f(c)$.

Step 1 — Substitute $x=c+h$.

As $x\to c$, the increment $h=x-c\to 0$. So we study $\displaystyle\lim_{h\to0}\sin(c+h)$ instead.

Step 2 — Expand using the sine addition formula.

Step 3 — Take the limit as $h\to0$.

Using the two standard results $\displaystyle\lim_{h\to0}\sin h = 0$ and $\displaystyle\lim_{h\to0}\cos h = 1$:

Step 4 — Compare with $f(c)$.

Since $f(c)=\sin c$, we get

All three continuity conditions ($f(c)$ defined, the limit exists, and the limit equals $f(c)$) hold.

Step 5 — Conclude for every point.

Because $c$ was an arbitrary real number, $f(x)=\sin x$ is continuous at every $c\in\mathbb{R}$ — that is, $\sin x$ is continuous on all of $\mathbb{R}$.

Method: Proving a Standard Function Is Continuous via an Inequality Bound (Epsilon-Delta Shortcut)

This method applies to functions like $\sin x$ where a direct algebraic identity lets you bound the change in output by the change in input, turning the epsilon-delta definition into a one-line argument.

Steps

Step 1: Write the difference $f(x) - f(a)$ using a known identity that separates it into a bounded factor and a "small" factor.

For sine, the sum-to-product identity gives:

Step 2: Bound the part that doesn't shrink.

Identify the factor whose magnitude is always at most a fixed constant (here $\left|\cos\left(\tfrac{x+a}{2}\right)\right| \leq 1$), so it can never amplify the difference.

Step 3: Use the standard inequality $|\sin\theta| \le |\theta|$ to bound the remaining factor.

This converts a trigonometric quantity into a simple algebraic one:

Step 4: Combine the bounds into a single clean inequality relating output-change to input-change.

Step 5: Finish the epsilon-delta argument.

Given any $\epsilon>0$, choosing $\delta=\epsilon$ (since the inequality is already this clean) guarantees $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. Because $a$ was arbitrary, this proves continuity at every real number.

Common Mistakes

Mistake 1: Assuming boundedness of a function (like $-1 \le \sin x \le 1$) by itself implies continuity.

Why it's wrong: many bounded functions are not continuous — a step function is bounded but jumps abruptly. Boundedness controls the range, not how the output responds to small changes in input, which is what continuity is actually about. Correct approach: always establish the input-to-output control (an inequality like $|f(x)-f(a)|\le|x-a|$) rather than citing boundedness alone.

Mistake 2: Forgetting to bound the cosine factor before using the sine inequality.

Why it's wrong: without the $|\cos(\cdot)|\le 1$ bound, the product $2\cos(\cdot)\sin(\cdot)$ can't be reduced to a clean single-variable inequality — skipping this step leaves the proof incomplete. Correct approach: explicitly state both bounds (cosine $\le 1$ and $|\sin\theta|\le|\theta|$) before multiplying them together.

Mistake 3: Choosing $\delta$ without deriving it from the actual inequality obtained.

Why it's wrong: $\delta$ must be chosen so that the derived inequality actually forces $|f(x)-f(a)|<\epsilon$ — picking an arbitrary $\delta$ without justification breaks the logical chain the epsilon-delta definition demands. Correct approach: only claim $\delta=\epsilon$ works after showing $|x-a|<\epsilon \Rightarrow |f(x)-f(a)| \le |x-a| < \epsilon$ explicitly.