Q.Show that $f(x) = |x - 5|$ is continuous but not differentiable at $x = 5$.

Concept: Differentiability of Absolute Value — the absolute value function has a sharp corner at the point where its argument is zero, making it non-differentiable there.

1. Continuity at $x=5$:

   $\lim_{x \to 5} |x-5| = 0$ and $f(5)=0$, so the limit equals the function value. Hence $f$ is continuous at $x=5$.

2. Left-hand derivative:

   For $x<5$, $f(x)=5-x$, so $f'(5^-) = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$.

3. Right-hand derivative: …

Why absolute value functions behave this way

The absolute value function $|x|$ has a V-shaped graph. At the vertex (where the expression inside becomes zero), the slope changes abruptly from $-1$ to $+1$. For $f(x) = |x-5|$, that vertex is shifted to $x=5$. Continuity is about the graph being unbroken — and the V is unbroken. Differentiability is about having a unique tangent — and at the tip of the V, there are infinitely many lines that touch the graph, not a single tangent.

Let’s verify both properties formally.

Step-by-step verification

1. Rewrite the function piecewise

The definition of absolute value gives:

So $f(x)$ is a straight line of slope $+1$ for $x \geq 5$, and slope $-1$ for $x < 5$.

2. Check continuity at $x=5$

A function is continuous at a point if three things match: the left-hand limit, the right-hand limit, and the function value.

• Left-hand limit ($x \to 5^-$):

  For $x<5$, $f(x)=5-x$.

  $\displaystyle \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (5-x) = 5-5 = 0$.

• Right-hand limit ($x \to 5^+$):

  For $x>5$, $f(x)=x-5$.

  $\displaystyle \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x-5) = 5-5 = 0$.

• Function value:

  $f(5) = |5-5| = 0$.

Since $\displaystyle \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5) = 0$, the function is continuous at $x=5$.

3. Check differentiability at $x=5$

Differentiability requires the derivative from the left and the derivative from the right to be equal. We compute each using the limit definition of the derivative.

Left-hand derivative (approach from $x<5$):

For $h<0$, $5+h < 5$, so $f(5+h) = 5 - (5+h) = -h$. And $f(5)=0$.

Right-hand derivative (approach from $x>5$): …

Method: Piecewise Continuity and Differentiability Check for $|x-a|$-Type Functions

This method applies whenever a function is defined using an absolute value (or any expression with a "kink"), and you must show it is continuous at the kink point but decide separately whether it is differentiable there.

Steps

Step 1: Rewrite the function piecewise around the critical point

Identify the point $x=a$ where the expression inside the modulus becomes zero (here $a=5$), and split $f(x)=|x-a|$ into two linear pieces — one for $x \geq a$ and one for $x < a$ — since $|x-a|=x-a$ when $x\geq a$ and $|x-a|=a-x$ when $x<a$.

Step 2: Test continuity using the three-way match

A function is continuous at $x=a$ only if the left-hand limit, the right-hand limit, and the function value all agree:

For a $|x-a|$-type function this always holds, because both pieces meet at height $0$ at $x=a$ — the graph has no break, only a corner.

Step 3: Test differentiability using one-sided derivatives

Continuity is necessary but never sufficient for differentiability. Compute the left-hand and right-hand derivatives separately using the limit definition: …

Common Mistakes

Mistake 1: Assuming continuity guarantees differentiability

Why it's wrong: continuity only requires the graph to have no break or jump; differentiability additionally requires a single, well-defined tangent slope. A V-shaped graph like $|x-5|$ is perfectly unbroken at the corner, so it is continuous, but the slope jumps from $-1$ to $+1$ there, so no single tangent exists. Correct approach: always check continuity and differentiability as two separate conditions — never infer one from the other.

Mistake 2: Differentiating $|x-5|$ as if it behaves like $x-5$ everywhere …