Q.Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

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

1. The function is $f(x) = |x|$. At $x = 0$, $f(0) = |0| = 0$.

2. For $x \to 0$, consider the left-hand limit: $\lim_{x \to 0^-} |x| = \lim_{x \to 0^-} (-x) = 0$.

   The right-hand limit: $\lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0$.

3. Since both one-sided limits equal $0$, $\lim_{x \to 0} |x| = 0$, which matches $f(0)$.

Why This Question Matters

Many students see the V-shaped graph of $|x|$ with its sharp point at $0$ and instinctively think "that's not smooth, so it must be discontinuous." That instinct confuses differentiability with continuity. A function can be perfectly continuous at a point even if it has a corner there — continuity only cares about whether the graph is unbroken, not whether it's smooth.

The definition of continuity at a point $x = a$ has three requirements, all of which must hold:

In plain language: as you approach $a$ from either side, the function values must settle down to the same number, and that number must be exactly what the function spits out at $a$.

Step-by-Step Verification

1. Write the function in piecewise form.

The absolute value function is defined differently for negative and non-negative inputs:

This piecewise form makes limits easy to compute — each piece is just a straight line.

2. Compute the left-hand limit as $x \to 0^-$.

When $x$ is just less than $0$, we use the top rule $f(x) = -x$. As $x$ gets arbitrarily close to $0$ from the left, $-x$ gets arbitrarily close to $0$:

3. Compute the right-hand limit as $x \to 0^+$.

When $x$ is just greater than $0$, we use the bottom rule $f(x) = x$. As $x$ approaches $0$ from the right, $x$ itself approaches $0$:

4. Compare the two one-sided limits.

Both are $0$, so the two-sided limit exists and equals $0$:

5. Evaluate the function at $x = 0$.

From the piecewise definition, at $x = 0$ we use the second rule: $f(0) = 0$.

6. Check the continuity condition.

We have $\lim_{x \to 0} f(x) = 0$ and $f(0) = 0$. Since they match, all three conditions are satisfied.

Method: Testing Continuity at a Specified Point (Three-Condition Test)

This method checks whether a function is continuous at one named point by directly applying the three-part continuity definition, using one-sided limits whenever the formula changes near that point.

Steps

Step 1: Evaluate the function at the given point

Compute $f(a)$ directly from the function's definition. If $f(a)$ is undefined, the function cannot be continuous at $a$ and the test stops here.

Step 2: Rewrite the function near the point, if needed

If the function involves an absolute value, a modulus, or any expression whose formula changes near $a$, rewrite it in piecewise form valid just to the left and just to the right of $a$.

Step 3: Compute the left-hand and right-hand limits

Use the appropriate piece of the function for each side. If the two one-sided limits are unequal, the two-sided limit does not exist and the function is discontinuous at $a$ — stop here.

Step 4: Compare the common limit with the function value

If both one-sided limits agree on a common value $L$, check whether $L = f(a)$. If they are equal, all three continuity conditions hold and the function is continuous at $a$; if not, it is discontinuous there even though the limit exists.

Common Mistakes

Mistake 1: Assuming a sharp corner in the graph means the function is discontinuous

Why it's wrong: continuity only requires the graph to be unbroken — it says nothing about smoothness. $|x|$ has a corner at $x=0$ but the left-hand limit, right-hand limit, and function value all still agree there. Correct approach: check the three continuity conditions explicitly rather than judging by the shape of the graph; save the smooth-vs-sharp question for differentiability, a separate and stricter property.

Mistake 2: Writing $|x| = x$ for all $x$ without splitting into cases

Why it's wrong: $|x|$ equals $x$ only when $x \ge 0$ and equals $-x$ when $x < 0$; using a single formula near $x=0$ makes it impossible to compute the correct one-sided limits. Correct approach: always rewrite $|x|$ in piecewise form before taking the left-hand and right-hand limits at $x=0$.