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

The Continuity Condition: When a Function Has No "Breaks"

If you can trace a curve without ever lifting your pen — no jumps, gaps, or leaps — that curve is continuous. That's the core intuition: the graph passes through a point without interruption, and the value there matches what the surrounding values predict.

The Intuition: Three Things Must Align

For $f(x)$ to be continuous at $x = a$, three things must hold:

1. $f$ is defined at $a$ — there is a point $(a, f(a))$.
2. $f$ approaches a single value as $x \to a$ — the left and right sides agree.
3. That value equals $f(a)$ — no "hole" with a different value plugged in.

If any of these fails, $f$ is discontinuous at $a$.

The Precise Statement

$f$ is continuous at $x = a$ if and only if:

That one equation packs all three conditions: the limit exists (left and right limits equal and finite), $f(a)$ is defined, and they are equal. If $f$ is continuous at every point of $(a, b)$, it is continuous on that interval.

Common Pitfalls

The "hole" mistake: $f(x) = \frac{x^2 - 1}{x - 1}$ is undefined at $x = 1$. Even though $\lim_{x \to 1} f(x) = 2$ exists, $f(1)$ doesn't — discontinuous.

The "jump" mistake: piecewise functions often cause this. For

at $x = 2$ the left limit is $3$, the right limit is $4$ — they don't match, so the limit doesn't exist.

The "blow-up" mistake: $f(x) = \frac{1}{x}$ at $x = 0$ is undefined and the limit goes to $\pm\infty$ — discontinuous.

Why It Matters

Continuity is the foundation for calculus. Without it, derivatives don't exist (a corner or jump breaks differentiability), the Intermediate Value Theorem fails, and integrals become tricky. …