Q.For what value of $\lambda$ is the function defined by $f(x) = \begin{cases} \lambda(x^2-2x), & \text{if } x \le 0 \\ 4x+1, & \text{if } x > 0 \end{cases}$ continuous at $x=0$? What about continuity at $x=1$?

Concept: Continuity At A Point — a function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: Continuity at $x = 0$

For $x \le 0$, $f(0) = \lambda(0^2 - 0) = 0$.

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

Right-hand limit: $\lim_{x \to 0^+} f(x) = 4(0) + 1 = 1$.

Step 2: For continuity at $0$, we need $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

Here $0 \neq 1$, so no value of $\lambda$ can make the limits equal.

Step 3: Continuity at $x = 1$

At $x = 1$, only the second piece applies ($x > 0$). …

The core idea: continuity at a point

A function $f$ is continuous at $x = a$ if three things match perfectly:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists (both sides agree).
3. That limit equals $f(a)$.

For a piecewise function, the danger zone is always the boundary between the two pieces. At $x=0$, the definition switches from $\lambda(x^2 - 2x)$ (for $x \le 0$) to $4x+1$ (for $x > 0$). So we must check the left-hand limit, the right-hand limit, and the value at $0$ itself.

Step-by-step work

1. Find $f(0)$

Since $x=0$ falls in the first piece ($x \le 0$), we use $f(x) = \lambda(x^2 - 2x)$:

So $f(0) = 0$ for any $\lambda$. That's fine — the function is defined at $0$ no matter what.

2. Left-hand limit as $x \to 0^-$

For $x$ just less than $0$, we are still in the first piece:

Both $x^2$ and $-2x$ are continuous, so we can substitute directly:

So the left-hand limit is $0$, regardless of $\lambda$.

3. Right-hand limit as $x \to 0^+$

For $x$ just greater than $0$, we use the second piece:

Again, substitute directly:

So the right-hand limit is $1$, and it does not depend on $\lambda$ at all.

4. The clash at $x=0$

For continuity at $0$, we need:

We have:

• Left-hand limit = $0$
• Right-hand limit = $1$
• $f(0) = 0$

The left-hand limit equals $f(0)$, but the right-hand limit is $1$, not $0$. No matter what $\lambda$ we pick, the right-hand limit stays $1$. So the two one-sided limits never agree — the overall limit does not exist. …

Method: Testing Continuity of a Piecewise Function Containing an Unknown Parameter

This method applies whenever a piecewise function has an unknown constant (like $\lambda$ or $k$) and you must find the value(s) — if any — that make it continuous at the join point, and separately check continuity at points away from the join.

Steps

Step 1: Identify the join point(s)

A piecewise function can only be discontinuous where its defining rule changes. Locate every boundary $x=a$ where the formula switches; every other point uses a single, standard (polynomial/trig/etc.) expression and is automatically continuous there, since such expressions are continuous on their whole domain.

Step 2: Compute the function value using the piece that actually covers the boundary

Check which inequality (strict or non-strict) includes the boundary point itself, and use exactly that piece to compute $f(a)$. This is the most common place students slip up — see the Common Mistakes tab.

Step 3: Compute the left-hand and right-hand limits separately

using the piece valid on each side. If the piece is a standard continuous function, these limits are found by direct substitution.

Step 4: Set up and solve the continuity equation

For continuity at $a$, require …

Common Mistakes

Mistake 1: Concluding "no value of $\lambda$ works" too hastily without checking which piece actually gives $f(0)$

A student might try to solve $\lambda(x^2-2x)=4x+1$ directly at $x=0$ without separately computing the left-hand limit, right-hand limit, and $f(0)$ as three distinct quantities. Why it's wrong: continuity needs all three to agree, not just the boundary equation to "balance." Correct approach: compute $f(0)$, the LHL, and the RHL independently, then compare.

Mistake 2: Assuming continuity failing at $x=0$ means the function is discontinuous everywhere, including at $x=1$

Why it's wrong: continuity is a pointwise property — a function can fail at one point and still be perfectly continuous elsewhere. Correct approach: check $x=1$ completely independently, using only the piece ($4x+1$) that applies in a neighbourhood of $1$.

Mistake 3: Forgetting that $\lambda \cdot 0 = 0$ regardless of $\lambda$'s value …