Q.Discuss the continuity of the function $f$, where $f$ is defined by $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$

For a piecewise function, check the boundary points where the formula changes — here $x=-1$ and $x=1$; elsewhere each piece is a polynomial and is continuous.

At $x=-1$: LHL $=\lim_{x\to -1^-}(-2)=-2$; RHL $=\lim_{x\to -1^+}2x=-2$; and $f(-1)=-2$. All equal, so $f$ is continuous here. …

A piecewise function can only break where its definition switches — here at $x=-1$ and $x=1$. On each open interval $(-\infty,-1)$, $(-1,1)$, $(1,\infty)$ the function is a constant or the line $2x$, all continuous. So we only need to test the two boundaries, checking $\lim_{x\to a^-}f=\lim_{x\to a^+}f=f(a)$.

At $x=-1$

Left piece ($x\le -1$) gives $-2$: $\displaystyle\lim_{x\to -1^-}f(x)=-2$.

Middle piece ($-1<x\le 1$) gives $2x$: $\displaystyle\lim_{x\to -1^+}2x=2(-1)=-2$.

Value: $f(-1)=-2$ (the first piece includes $x=-1$).

All three equal $-2$, so $f$ is continuous at $x=-1$.

At $x=1$

Middle piece gives $2x$: $\displaystyle\lim_{x\to 1^-}2x=2$. …

Method: Confirming Continuity Across Several Boundaries Without Assuming an Outcome

A function with several pieces is not automatically more likely to be discontinuous just because it has more switch points — this method shows how to verify (rather than guess) that every boundary actually holds up.

Steps

Step 1: Locate every switch point and note which piece owns each one

For each boundary, check the inequality symbols carefully to see which of the two neighbouring pieces includes the boundary value itself (the one with $\leq$ or $\geq$).

Step 2: Apply the three-condition test at each boundary in turn

Compute all three quantities from their respective pieces — never assume the outcome in advance, even if a boundary "looks" like it should behave a certain way.

Step 3: Treat a coincidental numeric match with care, not as automatic proof …

Common Mistakes

Mistake 1: Using the wrong neighbouring piece to compute $f(-1)$, even though the numbers happen to coincide

Why it's wrong: the boundary condition is $x \leq -1$, so $f(-1)$ must come from the constant piece ($-2$), not from the middle piece $2x$ — here both happen to equal $-2$ at $x=-1$, which can mask a genuine piece-selection error that would matter on a different problem. Correct approach: always confirm which inequality includes the equals sign before deciding which formula defines $f(a)$, even if the numbers seem to work out either way.

Mistake 2: Assuming a three-piece function with two switch points is more likely to contain a discontinuity than a simpler function …