Find the points at which $f$ is discontinuous. At which of these points is $f$ continuous from the right, from the left, or neither? Sketch the graph of $f$. **(i)** $f(x)=\begin{cases}2x+1, & x\le-1\\ 3x, & -1<x<1\\ 2x-1, & x\ge1\end{cases}$ **(ii)** $f(x)=\begin{cases}(x-1)^3, & x<0\\ (x+1)^3, & x\ge0\end{cases}$

Find the points at which f is discontinuous. At which of these points is f continuous from the right, from the

Exercise 9.5 · Q9

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Concept understanding — Continuity of a Function

Continuity of a Function

The Intuition: Drawing Without Lifting the Pen

Imagine you are drawing the graph of a function on a piece of paper. If you can trace the entire curve without lifting your pen from the paper, the function is continuous. Every time you have to lift the pen — because the graph jumps, breaks, or has a hole — the function is discontinuous at that point.

That is the visual idea. A continuous function has no sudden leaps, no gaps, no punctures. Its output changes smoothly as its input changes.

Consider a simple example: $f(x) = x^2$ . As $x$ moves from 1 to 2, the output moves from 1 to 4, passing through every value in between. No jump, no missing point. You can draw it in one stroke.

Now contrast that with a function like:

f(x) = \begin{cases} x^2 & \text{if } x \neq 1 \\ 5 & \text{if } x = 1 \end{cases}

At $x=1$ , the graph has a single isolated point at height 5, while the rest of the curve approaches height 1. To draw this, you would trace the parabola, then lift your pen to place a dot at $(1,5)$ . That lift is the discontinuity.

The Problem with Intuition Alone

"Drawing without lifting the pen" works for simple functions, but it fails for strange ones. Some functions are continuous yet impossible to draw (like the Weierstrass function, which is continuous but has no smooth tangent anywhere). More practically, the pen-lifting test is not a mathematical definition — it cannot tell you exactly what "no break" means at a single point.

We need a precise, point-by-point definition.

The Precise Definition: The Three-Part Test

A function $f(x)$ is said to be continuous at a point $x = a$ if and only if all three of the following conditions hold:

$f(a)$ is defined. The function must have a value at $x = a$ . No holes.
$\lim_{x \to a} f(x)$ exists. As $x$ gets arbitrarily close to $a$ from either side, the function's values must approach a single finite number.
$\lim_{x \to a} f(x) = f(a)$ . The limit must equal the actual function value. The point must sit exactly where the surrounding curve is heading.

If any one of these fails, the function is discontinuous at $x = a$ .

Important

Condition 3 is the heart of continuity. It says: "What the function should be (the limit) is exactly what it is (the value)." No surprises.

Why the Limit Matters

The limit captures the trend of the function near $a$ , ignoring what happens exactly at $a$ . Continuity demands that this trend matches the actual point. This is why the earlier piecewise function fails: the limit as $x \to 1$ is $1$ , but $f(1) = 5$ , so condition 3 is violated.

A Worked Example

Test whether $f(x) = \frac{x^2 - 1}{x - 1}$ is continuous at $x = 1$ .

Step 1: Is $f(1)$ defined? …

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For each join point, compute $\lim_{x\to x_0^-}f(x)$ , $\lim_{x\to x_0^+}f(x)$ , and $f(x_0)$ ; the point is a jump discontinuity whenever the two one-sided limits both exist but differ, and it is "continuous from the side" whose one-sided limit matches $f(x_0)$ .

Step 1. Part (i) — join at $x=-1$ . $f(x)=2x+1$ for $x\le-1$ , $f(x)=3x$ for $-1<x<1$ . Left-hand limit: $\lim_{x\to-1^-}(2x+1)=-2+1=-1$ . Since $x\le-1$ includes $-1$ , $f(-1)=2(-1)+1=-1$ . Right-hand limit: $\lim_{x\to-1^+}3x=-3$ . Left limit $=f(-1)=-1$ but right limit $=-3\ne-1$ : the one-sided limits disagree, so $f$ is discontinuous at $x=-1$ (a jump discontinuity), and since the left-hand limit matches $f(-1)$ , $f$ is continuous from the left (but not from the right) at $x=-1$ .

Step 2. Part (i) — join at $x=1$ . $f(x)=3x$ for $-1<x<1$ , $f(x)=2x-1$ for $x\ge1$ . Left-hand limit: $\lim_{x\to1^-}3x=3$ . Since $x\ge1$ includes $1$ , $f(1)=2(1)-1=1$ . Right-hand limit: $\lim_{x\to1^+}(2x-1)=1$ . Right limit $=f(1)=1$ but left limit $=3\ne1$ : another jump discontinuity, and since the right-hand limit matches $f(1)$ , $f$ is continuous from the right (but not from the left) at $x=1$ .

Step 3. Part (i) — sketch and summary. The graph is the line $y=2x+1$ for $x\le-1$ (ending at the filled point $(-1,-1)$ ), then the line $y=3x$ on $(-1,1)$ (open circles at both ends, at $(-1,-3)$ and $(1,3)$ ), then the line $y=2x-1$ for $x\ge1$ (starting at the filled point $(1,1)$ ) — two visible jumps. $f$ is continuous everywhere except at $x=-1$ (left-continuous) and $x=1$ (right-continuous); each linear piece is continuous on its open sub-interval. …

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Concept understanding — Continuity of a Function

Continuity of a Function

The Intuition: Drawing Without Lifting the Pen

The Problem with Intuition Alone

The Precise Definition: The Three-Part Test

Why the Limit Matters

A Worked Example

Step 1 — factor the numerator

Step 2 — take the limit at $x=5$

Step 3 — condition for continuity

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