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

Concept: Continuity At A Point — For $f$ to be continuous at $x=0$, we need $\lim_{x\to 0} f(x) = f(0) = \frac12$.

Step 1: Write the limit for $x \ne 0$:

Step 2: Use the standard limit $\lim_{t\to 0} \frac{1 - \cos t}{t^2} = \frac12$. Rewrite:

Step 3: Take the limit as $x\to 0$:

$\lim_{x\to 0} \frac{1 - \cos kx}{(kx)^2} = \frac12, \quad \lim_{x\to 0} \frac{x}{\sin x} = 1$ …

The idea is simple: a function is continuous at a point if the value it takes there matches what the surrounding behaviour predicts. Here, $f(0)$ is given as $\frac12$, so we need the limit of $\frac{1-\cos kx}{x\sin x}$ as $x$ approaches $0$ to also be $\frac12$. The trick is to rewrite the expression so that we can use two fundamental trigonometric limits.

1. Set up the continuity condition. For $f$ to be continuous at $x=0$, we require

Since for $x \ne 0$, $f(x) = \dfrac{1-\cos kx}{x\sin x}$, we need

2. Rewrite using the half-angle identity. A standard trick: $1-\cos\theta = 2\sin^2(\theta/2)$. So

This turns the limit into

3. Separate into known limit forms. Write it as

Now multiply numerator and denominator strategically:

The first factor $\frac{\sin^2(kx/2)}{(kx/2)^2}$ is $\left( \frac{\sin(kx/2)}{kx/2} \right)^2$, whose limit as $x\to 0$ is $1^2 = 1$ (since $\lim_{t\to 0} \frac{\sin t}{t}=1$).

4. Simplify the remaining algebraic part. We are left with $2 \cdot 1 \cdot \lim_{x\to 0} \frac{(kx/2)^2}{x\sin x} = 2 \cdot \lim_{x\to 0} \frac{k^2 x^2 / 4}{x\sin x} = 2 \cdot \frac{k^2}{4} \cdot \lim_{x\to 0} \frac{x}{\sin x}.$ …

Method: Solving for a Parameter Using Standard Trigonometric Limits

This method applies to any continuity/limit problem where the expression involves $1-\cos(\cdot)$ or $\sin(\cdot)$ with an unknown constant inside the argument, and a limit must be matched to a given function value.

Steps

Step 1: Set up the continuity condition

Write down what must hold: the limit of the $x \ne 0$ (or $x \ne a$) expression as $x$ approaches the point must equal the function's defined value there.

Step 2: Rewrite using the standard limits

Two standard results handle almost every trigonometric limit of this type:

$\lim_{t \to 0} \frac{\sin t}{t} = 1, \qquad \lim_{t \to 0} \frac{1 - \cos t}{t^2} = \frac{1}{2}$ …

Common Mistakes

Mistake 1: Applying $\lim \frac{\sin t}{t}=1$ without matching the argument

Why it's wrong: the standard limit only equals $1$ when the expression inside the trig function is identical to the denominator's variable — using $\lim \frac{\sin kx}{x}=1$ directly (instead of correctly accounting for $k$) ignores the scaling factor introduced by the argument. Correct approach: always rewrite so the argument inside sine/cosine and the denominator match exactly, introducing a compensating factor outside.

Mistake 2: Dropping the $\pm$ when solving a squared equation for the constant …