Q.Examine the continuity of $f$, where $f$ is defined by $f(x) = \begin{cases} \sin x - \cos x, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$

The key idea is continuity at a point: a function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: Compute the limit as $x \to 0$ for $x \neq 0$.

Since $\sin x$ and $\cos x$ are continuous everywhere,

Step 2: The function value at $x = 0$ is given directly: $f(0) = -1$. …

The core idea here is continuity at a point. A function is continuous at $x = a$ if three things hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

For this piecewise function, the only potential trouble spot is $x = 0$, because that's where the definition changes. Everywhere else, $f(x) = \sin x - \cos x$ is a combination of continuous functions (sine and cosine), so it's automatically continuous. The question is whether the "patch" at $x = 0$ matches the behavior of the formula around it.

Let's check step by step.

1. Check $f(0)$ is defined.

   The problem explicitly gives $f(0) = -1$. So condition 1 is satisfied.

2. Find $\lim_{x \to 0} f(x)$.

   For $x \neq 0$, $f(x) = \sin x - \cos x$. We need the limit of this expression as $x$ approaches 0.

   Both $\sin x$ and $\cos x$ are continuous everywhere, so we can directly substitute $x = 0$:

The limit exists and equals $-1$.

Method: Checking Whether an Assigned Value at a Single Point Preserves Continuity

Use this when a function equals a standard continuous expression everywhere except at one specific point, where a particular numeric value is assigned by definition — the question is whether that assigned value happens to match what continuity would require.

Steps

Step 1: Compute the limit of the "everywhere else" expression as $x$ approaches the special point

Since that expression is built from standard continuous functions, evaluate the limit by direct substitution.

Step 2: Read off the assigned function value at the special point

This is given directly in the piecewise definition — no computation needed, just identify it.

Step 3: Compare the limit (Step 1) with the assigned value (Step 2)

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

Common Mistakes

Mistake 1: Assuming the piecewise definition automatically creates a discontinuity at $x=0$

Why it's wrong: a function can be defined "specially" at one point and still turn out continuous, if the assigned value happens to equal the limit of the surrounding formula — the mere presence of a separate case is not proof of a break. Correct approach: always compute $\lim_{x\to 0}(\sin x - \cos x)$ independently and compare it with the given $f(0)=-1$, rather than assuming the answer from the structure alone.

Mistake 2: Forgetting to substitute $\cos 0 = 1$ correctly (sign error) …