Q.Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} x^{10}-1, & \text{if } x \leq 1 \\ x^2, & \text{if } x > 1 \end{cases}$

Concept: Continuity At A Point — a function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For a piecewise function, check the left-hand limit, right-hand limit, and the function value at the break point.

Step 1: Evaluate $f(1)$

Since $x \leq 1$ uses the first piece,

$f(1) = 1^{10} - 1 = 0$.

Step 2: Compute left-hand limit as $x \to 1^-$

Using the same piece:

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1^{10} - 1 = 0$.

Step 3: Compute right-hand limit as $x \to 1^+$

Using the second piece:

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1$.

Step 4: Compare …

The function is

1. Away from $x = 1$. For $x < 1$, $f(x) = x^{10} - 1$ is a polynomial and hence continuous. For $x > 1$, $f(x) = x^2$ is a polynomial and hence continuous. So $f$ is continuous at every point except possibly $x = 1$.

2. At $x = 1$. From the definition ($x \leq 1$ branch), $f(1) = 1^{10} - 1 = 0$.

• Left-hand limit: $\displaystyle\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1 - 1 = 0.$ …

Method: Numerically Verifying Whether Two Polynomial Pieces Agree at Their Boundary

Even when both pieces of a function look like "nice," smooth polynomials, they can still disagree at the point where they meet — this method shows how to check that rigorously rather than by eye.

Steps

Step 1: Pin down $f(a)$ using the correctly-matched piece

Match the boundary value $a$ to whichever inequality includes equality, and substitute into that formula only.

Step 2: Substitute directly into each side's formula to get both one-sided limits

Direct substitution is valid here because polynomials have no gaps or jumps on their own open interval.

Step 3: Compare the two numeric results, not the formulas themselves …

Common Mistakes

Mistake 1: Assuming two smooth polynomial pieces automatically meet at the same height at their boundary

Why it's wrong: both $x^{10}-1$ and $x^2$ are well-behaved polynomials, but being individually smooth says nothing about whether they agree at the shared point $x=1$ — here they genuinely give different values ($0$ versus $1$). Correct approach: always substitute and compute the actual numbers on both sides rather than assuming agreement from how "nice" each piece looks. …