Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = |x| + |x - 1|$ at $x = 1$.

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

Step 1: Evaluate $f(1)$

$f(1) = |1| + |1 - 1| = 1 + 0 = 1$.

Step 2: Compute the left-hand limit ($x \to 1^-$)

For $x < 1$, $|x| = x$ and $|x-1| = -(x-1) = 1-x$.

So $f(x) = x + (1-x) = 1$. Hence $\lim_{x \to 1^-} f(x) = 1$.

Step 3: Compute the right-hand limit ($x \to 1^+$)

For $x > 1$, $|x| = x$ and $|x-1| = x-1$. …

Why This Approach Works

Continuity at a point is a local idea — it asks whether the function's value at that point matches what we'd expect from nearby values. For $f(x) = |x| + |x-1|$, the trouble spot is $x=1$ because the second absolute value term $|x-1|$ changes its behaviour there. Absolute value functions are piecewise linear, so the key is to rewrite $f$ without absolute values on either side of $x=1$, then compare the limits.

The definition is simple: $f$ is continuous at $x=a$ if $\lim_{x \to a} f(x) = f(a)$. For this to hold, both one-sided limits must exist and equal the function value.

Step-by-Step Solution

1. Understand the behaviour of $|x-1|$ near $x=1$

The absolute value $|x-1|$ is defined as:

• $|x-1| = x-1$ when $x \ge 1$
• $|x-1| = -(x-1) = 1-x$ when $x < 1$

The first term $|x|$ is simpler: $|x| = x$ for $x \ge 0$, and since we're near $x=1$, $x$ is positive, so $|x| = x$ on both sides.

2. Write the piecewise form of $f(x)$

For $x < 1$:

For $x \ge 1$:

So the function is:

3. Compute the left-hand limit as $x \to 1^-$

From the left, $f(x) = 1$ constantly. So:

4. Compute the right-hand limit as $x \to 1^+$

From the right, $f(x) = 2x - 1$. Substituting $x=1$:

5. Find the function value at $x=1$

Since $x=1$ falls in the $x \ge 1$ case:

6. Compare all three …

Method: Testing Continuity of a Sum of Absolute-Value Functions at a Critical Point

This method applies whenever the function is built from a sum (or combination) of absolute-value terms, and continuity must be checked exactly at a point where one of those terms changes its sign — the "corner" of that term.

Steps

Step 1: Identify every critical point where an individual absolute-value term changes sign.

For a term $|x-c|$, the sign change happens at $x=c$. List all such points; the point being tested for continuity will be one of them.

Step 2: Rewrite each absolute-value term without the bars, separately for $x$ below and above the critical point.

Use $|x-c| = x-c$ for $x \ge c$ and $|x-c| = -(x-c)$ for $x < c$, and do this for every term in the sum.

Step 3: Add the rewritten terms to get a single explicit (piecewise) formula for $f(x)$ on each side of the critical point, then evaluate the left-hand limit and right-hand limit from these simplified formulas. …

Common Mistakes

Mistake 1: Concluding the function is discontinuous at $x=1$ because $|x-1|$ has a sharp corner there.

Why it's wrong: a corner in the graph reflects a possible failure of differentiability, not continuity — the one-sided limits and the function value can still all agree at a corner, which is exactly what happens here (all equal $1$). Correct approach: judge continuity strictly by comparing the one-sided limits and $f(a)$, never by the smoothness (or lack of it) of the graph.

Mistake 2: Forgetting to also rewrite $|x|$ without bars, since near $x=1$ its sign never actually changes. …