Q.Find the value of $k$ so that the function $f$ is continuous at the indicated point: $f(x) = \begin{cases} \dfrac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}, & -1 \le x < 0 \\ \dfrac{2x + 1}{x - 1}, & 0 \le x \le 1 \end{cases}$ at $x = 0$.

Concept: Continuity At A Point — For $f$ to be continuous at $x=0$, the left-hand limit, right-hand limit, and $f(0)$ must all be equal.

Step 1: Compute $f(0)$

From the second piece, $f(0) = \frac{2(0)+1}{0-1} = -1$.

Step 2: Compute $\lim_{x \to 0^-} f(x)$

Rationalise the numerator: …

We need continuity at $x=0$. That means the function’s value at $0$ must match what the function approaches from both sides. The definition of $f$ changes at $x=0$: for $x<0$ we have one expression, for $x\ge0$ another. So we must compute the left-hand limit, the right-hand limit, and $f(0)$, then set them equal.

1. Find $f(0)$

Since $0$ falls in the second piece ($0 \le x \le 1$), we use $f(x) = \frac{2x+1}{x-1}$.

So the function’s value at the point is $-1$.

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

For $x>0$ (but near $0$), we again use the second piece. The expression is rational and continuous at $0$ (denominator $x-1$ is $-1$, not zero), so we can substitute directly:

So the right-hand limit is $-1$, which already matches $f(0)$.

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

For $x<0$, we use $f(x) = \frac{\sqrt{1+kx} - \sqrt{1-kx}}{x}$. Substituting $x=0$ directly gives $\frac{0}{0}$, an indeterminate form. So we need to simplify.

The classic trick: rationalise the numerator by multiplying numerator and denominator by the conjugate $\sqrt{1+kx} + \sqrt{1-kx}$.

The numerator becomes:

So the expression simplifies to:

$\frac{2kx}{x\left(\sqrt{1+kx} + \sqrt{1-kx}\right)} = \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}}, \quad x \neq 0.$ …

Method: Finding an Unknown Constant for Continuity of a Piecewise Function

This method applies whenever a piecewise function contains an unknown constant, and you're asked to choose its value so the function is continuous at the junction point.

Steps

Step 1: Identify which piece defines the function value at the junction point

A piecewise definition often includes the junction point itself in one of the branches (via $\le$ or $\ge$). Always read the piece's inequality carefully — plugging into the wrong branch is the single most common error here.

Step 2: Compute the one-sided limits from each branch …

Common Mistakes

Mistake 1: Using the wrong branch to evaluate $f(a)$ at the junction point

Why it's wrong: when the junction point itself is included in one specific piece (via the $\le$/$\ge$ in that branch's condition), some students plug it into the other, visually "nearby" branch instead. Correct approach: read the inequality on each branch carefully and match the junction point to the branch whose condition actually includes it.

Mistake 2: Leaving the limit in indeterminate $\frac{0}{0}$ form …