Q.Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} x+1, & \text{if } x \geq 1 \\ x^2+1, & \text{if } x < 1 \end{cases}$

The key idea is continuity at a point: a function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For a piecewise function, we check the left-hand limit, right-hand limit, and the function value at the boundary point $x = 1$.

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

Step 2: Compute the right-hand limit as $x \to 1^+$ (using $x + 1$):

Step 3: Evaluate $f(1)$ (using $x \geq 1$):

$f(1) = 1 + 1 = 2.$ …

The core idea of continuity at a point is simple: a function is continuous at $x = a$ if you can draw its graph near that point without lifting your pen. More formally, three things must hold:

1. The function is defined at $a$ (i.e., $f(a)$ exists).
2. The limit of $f(x)$ as $x$ approaches $a$ exists.
3. That limit equals $f(a)$.

For a piecewise function like this one, the only potential trouble spot is the boundary where the definition changes — here, $x = 1$. Everywhere else, the function is given by a polynomial ($x+1$ or $x^2+1$), and polynomials are continuous on their entire domain. So the entire question reduces to: What happens at $x = 1$?

Let’s work through it step by step.

1. Find $f(1)$. Since $x = 1$ satisfies $x \geq 1$, we use the top piece:

So the function is defined at $x=1$, and its value is $2$.

2. Compute the right-hand limit as $x \to 1^+$. For $x > 1$ (approaching from the right), the function is $f(x) = x+1$.

3. Compute the left-hand limit as $x \to 1^-$. For $x < 1$ (approaching from the left), the function is $f(x) = x^2 + 1$.

4. Compare the three values. We have:
   • $f(1) = 2$
   • $\displaystyle \lim_{x \to 1^+} f(x) = 2$
   • $\displaystyle \lim_{x \to 1^-} f(x) = 2$ …

Method: Testing Continuity at the Boundary of a Two-Piece Function

Use this method whenever a function is defined by exactly one formula on one side of a point and a different formula on the other side.

Steps

Step 1: Locate the single boundary point

Find the value of $x$ where the definition switches from one formula to the other — this is the only point that needs checking, since each formula is a polynomial (or other standard continuous expression) and is therefore automatically continuous on its own open interval.

Step 2: Evaluate the function's actual value at the boundary

Read the inequality signs carefully: only one of the two formulas actually "owns" the boundary point (the one whose inequality includes equality, $\leq$ or $\geq$). Use that formula to compute $f(a)$.

Step 3: Compute the two one-sided limits

$\lim_{x \to a^-} f(x) \ \text{using the formula valid just left of } a, \qquad \lim_{x \to a^+} f(x) \ \text{using the formula valid just right of } a$ …

Common Mistakes

Mistake 1: Assuming the function must be discontinuous because the two pieces are different types of expressions

Why it's wrong: here one piece is a line ($x+1$) and the other is a parabola ($x^2+1$) — a student may assume different-looking formulas can't possibly meet smoothly, but continuity depends only on whether the numeric values actually agree at the boundary, never on how similar the formulas look. Correct approach: always compute and compare the actual numbers, regardless of what the formulas look like.

Mistake 2: Using $x^2+1$ (the $x<1$ piece) to compute $f(1)$ itself …