Q.Show that the function $f(x) = |\sin x + \cos x|$ is continuous at $x = \pi$.

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

1. Evaluate $f(\pi)$:

   $f(\pi) = |\sin \pi + \cos \pi| = |0 + (-1)| = 1$.

2. Find the left-hand and right-hand limits. Since $|\cdot|$ and $\sin x + \cos x$ are continuous everywhere, the composition is continuous at $\pi$ — but we verify directly:

   $\lim_{x \to \pi} (\sin x + \cos x) = \sin \pi + \cos \pi = -1$, so

   $\lim_{x \to \pi} |\sin x + \cos x| = |-1| = 1$. …

Why This Approach Works

Continuity at a point is a local property — it tells us whether the function behaves nicely near that specific $x$-value. For $f(x) = |\sin x + \cos x|$ at $x = \pi$, the absolute value makes the function non-negative, but it doesn't introduce any jumps or breaks by itself. The real question is whether the expression inside the absolute value changes sign abruptly at $\pi$, which could cause a corner or a gap.

The standard test for continuity at $x = a$ is:

1. $f(a)$ must be defined.
2. $\lim_{x \to a} f(x)$ must exist (both one-sided limits equal).
3. The limit must equal $f(a)$.

We'll apply this step by step.

Step-by-Step Solution

1. Evaluate $f(\pi)$ directly.

Plug $x = \pi$ into the function:

So $f(\pi) = 1$ — condition (1) is satisfied.

2. Simplify the expression inside the absolute value.

A neat trick: $\sin x + \cos x$ can be written as a single sine wave. Multiply and divide by $\sqrt{2}$:

This identity holds because $\sin(A+B) = \sin A \cos B + \cos A \sin B$, and here $\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$.

So our function becomes:

3. Find the limit as $x \to \pi$.

We need $\lim_{x \to \pi} f(x)$. Since $f(x) = \sqrt{2} \, |\sin(x + \pi/4)|$, and the absolute value function is continuous everywhere, we can focus on the inner sine function.

At $x = \pi$, the argument of sine is:

Now, $\sin(5\pi/4) = -\frac{\sqrt{2}}{2}$. So:

But we need the limit, not just the value. Since $\sin$ is continuous everywhere, $\sin(x + \pi/4)$ is continuous, and the absolute value of a continuous function is also continuous. Therefore:

$\lim_{x \to \pi} f(x) = f(\pi) = 1.$ …

Method: Proving Continuity of an Absolute-Value-of-a-Trig-Expression at a Point

This method applies to functions of the form $f(x) = |g(x)|$, where $g$ is built from standard continuous pieces (sine, cosine, sums), and you must confirm continuity at a specific point.

Steps

Step 1: Evaluate the function directly at the given point

Substitute the point into the original expression, applying the absolute value last, to get $f(a)$.

Step 2: Use the composition-of-continuous-functions idea

Since $\sin x$, $\cos x$, their sum, and the absolute value function are all continuous everywhere, $|g(x)|$ is continuous wherever $g(x)$ itself is — which, for a combination of sine and cosine, is every real number. This alone certifies continuity without further limit computation.

Step 3: Verify explicitly by computing the two one-sided limits

Even though Step 2 already guarantees the result, an exam answer should still confirm it directly:

$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$ …

Common Mistakes

Mistake 1: Assuming the absolute value automatically threatens continuity

Why it's wrong: some students treat $|g(x)|$ as inherently risky and hunt for a "break" near the point, even when $g(x)$ never touches zero there — but the absolute value of any continuous function is itself continuous everywhere, so there is nothing extra to prove beyond confirming $g$ is continuous. Correct approach: recall that $|{\cdot}|$ is a continuous function, so a composition with a continuous $g$ is continuous by the composition rule — no special-casing is needed unless $g(a)=0$.

Mistake 2: Sign or evaluation errors substituting standard angle values …