Q.Discuss the continuity of the function $f$, where $f$ is defined by $f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \le x \le 1 \\ 4x, & \text{if } x > 1 \end{cases}$

Concept: Piecewise Function Continuity — check the function at the boundary points where the definition changes.

Step 1: At $x = 0$

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

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

Function value: $f(0) = 0$

Since LHL = RHL = $f(0)$, $f$ is continuous at $x = 0$.

Step 2: At $x = 1$

Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 0 = 0$

Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4x = 4$

Function value: $f(1) = 0$ …

The question asks us to discuss the continuity of a piecewise function. A piecewise function is defined by different expressions on different intervals. The only places where continuity can break are at the boundaries where the expression changes — here, at $x=0$ and $x=1$. Everywhere else, each piece is a simple polynomial ($2x$, $0$, or $4x$), and polynomials are continuous on open intervals. So the entire job reduces to checking those two points.

For a function to be continuous at a point $x = a$, three things must hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists (both one-sided limits must be equal).
3. $\lim_{x \to a} f(x) = f(a)$.

We'll apply this to $x=0$ and $x=1$ separately.

1. Check continuity at $x = 0$

The definition splits at $0$: for $x<0$, $f(x)=2x$; for $0 \le x \le 1$, $f(x)=0$.

• Left-hand limit as $x \to 0^-$:

  For $x$ just less than $0$, we use $f(x)=2x$.

  $\displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x = 2(0) = 0$.

• Right-hand limit as $x \to 0^+$:

  For $x$ just greater than $0$ (but still less than $1$), we use $f(x)=0$.

  $\displaystyle \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0$.

• Function value at $x=0$:

  Since $0$ falls in the interval $0 \le x \le 1$, $f(0) = 0$.

All three are $0$, so the limit exists and equals the function value.

Conclusion: $f$ is continuous at $x=0$.

2. Check continuity at $x = 1$

Here the definition splits again: for $0 \le x \le 1$, $f(x)=0$; for $x>1$, $f(x)=4x$.

• Left-hand limit as $x \to 1^-$:

  For $x$ just less than $1$ (but $\ge 0$), we use $f(x)=0$.

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

• Right-hand limit as $x \to 1^+$:

  For $x$ just greater than $1$, we use $f(x)=4x$. …

Method: Distinguishing "Continuous at This Boundary" From "Continuous Everywhere"

This method is for functions with multiple pieces where different boundaries can behave differently — some continuous, some not — within the same function.

Steps

Step 1: Identify all switch points

Note every $x$-value where the piecewise rule changes; these are the only candidates for discontinuity, since each individual piece is continuous on its own open interval.

Step 2: Test the first boundary on its own

Compute $f(a_1)$, the left-hand limit, and the right-hand limit at that boundary using only the pieces that meet there. Reach a verdict (continuous or discontinuous) for that boundary alone.

Step 3: Test the next boundary independently …

Common Mistakes

Mistake 1: Treating the function as either "fully continuous" or "fully discontinuous" instead of testing each boundary separately

Why it's wrong: this function passes the continuity test at $x=0$ but fails it at $x=1$ — a single overall verdict for the whole function hides this genuinely mixed result. Correct approach: report continuity or discontinuity separately for each boundary point, not as one blanket statement for the entire function.

Mistake 2: At $x=1$, noting $f(1)=0$ matches the left-hand limit and stopping there …