Q.If $f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \le x \le 1 \\ nx^3 + m, & x > 1 \end{cases}$. For what integers $m$ and $n$ does both $\lim_{x\to 0} f(x)$ and $\lim_{x\to 1} f(x)$ exist?

Concept: Limit Of A Piecewise Function — for the limit to exist at a boundary, the left-hand and right-hand limits must be equal.

Step 1: Limit at $x = 0$

Left-hand limit ($x \to 0^-$): $\lim_{x\to 0^-} f(x) = m(0)^2 + n = n$

Right-hand limit ($x \to 0^+$): $\lim_{x\to 0^+} f(x) = n(0) + m = m$

Equality requires $n = m$.

Step 2: Limit at $x = 1$

Left-hand limit ($x \to 1^-$): $\lim_{x\to 1^-} f(x) = n(1) + m = n + m$

Right-hand limit ($x \to 1^+$): $\lim_{x\to 1^+} f(x) = n(1)^3 + m = n + m$ …

The key idea is that a limit exists at a boundary point if and only if the function approaches the same value from both sides. For a piecewise function, this means the expressions on either side of the boundary must give the same result as $x$ approaches that point.

Let’s unpack why this works. A limit is about behaviour near a point, not at the point itself. So at $x=0$, we don’t care what $f(0)$ is (that’s $n$ from the middle piece) — we care what the left piece $mx^2+n$ approaches as $x\to 0^-$, and what the middle piece $nx+m$ approaches as $x\to 0^+$. If those two one-sided limits are equal, the two-sided limit exists.

Similarly at $x=1$, we compare the middle piece (as $x\to 1^-$) with the right piece (as $x\to 1^+$).

Now let’s work through it step by step.

1. At $x=0$:

   • Left-hand limit ($x\to 0^-$): $f(x) = mx^2 + n$. As $x\to 0$, $x^2\to 0$, so this approaches $m\cdot 0 + n = n$.
   • Right-hand limit ($x\to 0^+$): $f(x) = nx + m$. As $x\to 0$, this approaches $n\cdot 0 + m = m$.
   • For $\lim_{x\to 0} f(x)$ to exist, we need $n = m$.

2. At $x=1$:

   • Left-hand limit ($x\to 1^-$): $f(x) = nx + m$. As $x\to 1$, this approaches $n\cdot 1 + m = n + m$.
   • Right-hand limit ($x\to 1^+$): $f(x) = nx^3 + m$. As $x\to 1$, $x^3\to 1$, so this approaches $n\cdot 1 + m = n + m$.
   • The two sides are automatically equal — no condition needed. So $\lim_{x\to 1} f(x)$ exists for any integers $m,n$. …