Q.Find the values of $k$ so that the function $f$ is continuous at the indicated point, where $f$ is defined by $f(x) = \begin{cases} kx+1, & \text{if } x \leq \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ at $x = \pi$

Concept: Continuity Condition — For $f$ to be continuous at $x = \pi$, the left-hand limit, right-hand limit, and $f(\pi)$ must all be equal.

Step 1: Compute $f(\pi)$ and the left-hand limit. For $x \leq \pi$, $f(x) = kx + 1$, so

$f(\pi) = k\pi + 1$ and $\lim_{x \to \pi^-} f(x) = k\pi + 1$.

Step 2: Compute the right-hand limit. For $x > \pi$, $f(x) = \cos x$, so

$\lim_{x \to \pi^+} f(x) = \cos \pi = -1$.

Step 3: Set the two equal for continuity: …

We are checking continuity at the point where the definition of $f$ changes — $x = \pi$. The function is given by two different expressions on either side of this point. For continuity, the function must not "jump" when we cross $x = \pi$; the value coming from the left must match the value coming from the right, and both must match what the function actually gives at $x = \pi$.

The left-hand piece ($x \leq \pi$) gives $f(x) = kx + 1$, so at $x = \pi$ itself, the function is defined as $f(\pi) = k\pi + 1$. The right-hand piece ($x > \pi$) gives $f(x) = \cos x$, which does not include $x = \pi$ — but it tells us what values the function takes as we approach $\pi$ from the right.

Let’s work through the three conditions for continuity at $x = \pi$.

1. Find $f(\pi)$ directly. Since $\pi$ satisfies $x \leq \pi$, we use the first piece:

2. Compute the left-hand limit as $x \to \pi^-$. For $x$ just less than $\pi$, the function is still $kx + 1$. Since $kx + 1$ is a polynomial (continuous everywhere), the limit as $x$ approaches $\pi$ from the left is simply the value at $\pi$:

3. Compute the right-hand limit as $x \to \pi^+$. For $x$ just greater than $\pi$, the function is $\cos x$. The cosine function is continuous everywhere, so the limit as $x$ approaches $\pi$ from the right is:

4. Set the three equal for continuity. For $f$ to be continuous at $x = \pi$, we need: $\lim_{x \to \pi^-} f(x) = f(\pi) = \lim_{x \to \pi^+} f(x).$ …

Method: Finding an Unknown Constant from the Continuity Condition

This method applies whenever a piecewise function must be continuous at the single point where its rule switches, and one piece contains an unknown constant.

Steps

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

Look at the inequality signs in the piecewise definition. Whichever piece's inequality includes the junction point itself (a $\leq$ or $\geq$, not a strict $<$ or $>$) is the one that gives $f(a)$ directly.

Step 2: Write the left-hand limit using the piece valid just before the junction

Substitute the junction value into that piece's expression. If the piece is a polynomial or a standard continuous function (like a trigonometric function), the limit as $x$ approaches the junction equals the piece evaluated exactly at the junction.

Step 3: Write the right-hand limit using the piece valid just after the junction …

Common Mistakes

Mistake 1: Using the wrong piece to evaluate $f(\pi)$ itself

Why it's wrong: since the condition is $x \leq \pi$, the point $x=\pi$ belongs to the first piece ($kx+1$), not the second. A student who plugs $\pi$ into $\cos x$ to get $f(\pi)$ is evaluating a formula that isn't even defined at that point. Correct approach: always check which inequality includes the equals sign before deciding which expression gives the function's actual value at the junction.

Mistake 2: Sign error on $\cos \pi$ …