Q.Find the values of $k$ so that the function $f$ is continuous at the indicated point, where $f$ is defined by $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

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

Step 1: Evaluate $f(\pi/2)$

Given directly: $f(\pi/2) = 3$.

Step 2: Compute $\lim_{x \to \pi/2} f(x)$

For $x \neq \pi/2$, $f(x) = \frac{k \cos x}{\pi - 2x}$.

Let $t = \pi/2 - x$, so $x = \pi/2 - t$ and $t \to 0$. Then $\cos x = \cos(\pi/2 - t) = \sin t$, and $\pi - 2x = \pi - 2(\pi/2 - t) = 2t$.

Thus the limit becomes: …

The core idea here is continuity at a point. A function $f$ is continuous at $x = a$ if three things hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

In this problem, $f\!\left(\frac{\pi}{2}\right) = 3$ is given, so condition 1 is satisfied. The real work is checking whether the limit of $f(x)$ as $x$ approaches $\frac{\pi}{2}$ exists and equals 3. Since the function is defined piecewise, the limit depends on the expression $\frac{k \cos x}{\pi - 2x}$ for $x \neq \frac{\pi}{2}$.

The tricky part: as $x \to \frac{\pi}{2}$, both the numerator $\cos x$ and the denominator $\pi - 2x$ approach 0. This gives a $\frac{0}{0}$ indeterminate form. We need to resolve this limit to see what value it approaches, and then choose $k$ so that this limit matches 3.

1. Set up the limit we need to evaluate

   We want $\lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$.

   Since $k$ is a constant, it factors out: $k \cdot \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\pi - 2x}$.

2. Use a substitution to simplify the limit

   The denominator $\pi - 2x$ suggests letting $t = \frac{\pi}{2} - x$. Then as $x \to \frac{\pi}{2}$, we have $t \to 0$. Also, $x = \frac{\pi}{2} - t$.

   Substitute into the denominator: $\pi - 2x = \pi - 2\left(\frac{\pi}{2} - t\right) = \pi - \pi + 2t = 2t$.

   For the numerator: $\cos x = \cos\!\left(\frac{\pi}{2} - t\right) = \sin t$.

   So the limit becomes:

3. Apply the standard limit

   The fundamental trigonometric limit is $\lim_{t \to 0} \frac{\sin t}{t} = 1$. This is a result you should know by heart — it's the backbone of many continuity and derivative problems involving sine.

   Therefore:

   $\lim_{x \to \frac{\pi}{2}} f(x) = \frac{k}{2} \cdot 1 = \frac{k}{2}.$ …

Method: Finding an Unknown Constant by Resolving a $\frac{0}{0}$ Trigonometric Limit

Use this when a piecewise function's assigned value at a point depends on an unknown constant appearing inside a limit that evaluates to an indeterminate $\frac{0}{0}$ form at that point — typically resolved with a substitution that reduces it to the standard limit $\lim_{t\to0}\frac{\sin t}{t}=1$.

Steps

Step 1: Confirm the indeterminate form

Plug the target point directly into the "elsewhere" piece's numerator and denominator; if both go to $0$, direct substitution fails and a limit technique is needed (never invent a method outside the syllabus — this chapter's tools are substitution tricks and standard limits, not derivative-based shortcuts).

Step 2: Choose a substitution that centers the limit at $0$

Let $t$ measure the distance from $x$ to the target point (e.g. $t = a - x$), so that $t \to 0$ as $x \to a$. Rewrite every trig term in the numerator/denominator in terms of $t$, using standard co-function identities (like $\cos(\pi/2 - t) = \sin t$).

Step 3: Simplify until the standard limit form appears

$\lim_{t \to 0} \frac{\sin t}{t} = 1$ …

Common Mistakes

Mistake 1: Trying to evaluate the $\frac{0}{0}$ limit by direct substitution without a technique

Why it's wrong: plugging $x=\pi/2$ directly into $\dfrac{k\cos x}{\pi - 2x}$ gives $\dfrac{0}{0}$, which is indeterminate — it does NOT mean the limit is zero, one, or undefined; more work is required. Correct approach: use a substitution (like $t = \pi/2 - x$) to convert the expression into a recognizable standard-limit form before evaluating.

Mistake 2: Getting the substitution's trig identity wrong (mixing up $\cos(\pi/2 - t)$ and $\sin(\pi/2-t)$)

Why it's wrong: the co-function identities $\cos(\pi/2-t)=\sin t$ and $\sin(\pi/2-t)=\cos t$ are easy to swap under pressure, and swapping them changes which standard limit applies and flips the whole answer. Correct approach: verify the identity by checking a specific value (e.g. at $t=0$, $\cos(\pi/2)=0=\sin 0$ ✓) before trusting it in the general substitution. …