Q.Suppose $f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}$ and if $\lim_{x\to 1} f(x) = f(1)$ what are possible values of $a$ and $b$?

The function is continuous at $x = 1$ if the limit from both sides equals the function value at that point.

Concept: For $\lim_{x \to 1} f(x) = f(1)$, we need both one-sided limits to exist and equal $f(1) = 4$.

Left-hand limit: As $x \to 1^-$, we use $f(x) = a + bx$:

Right-hand limit: As $x \to 1^+$, we use $f(x) = b - ax$:

$\lim_{x \to 1^+} f(x) = b - a(1) = b - a$ …

The condition $\lim_{x \to 1} f(x) = f(1)$ is precisely the definition of continuity at $x = 1$. We already know that $f(1) = 4$ from the piecewise definition. The question is: what must $a$ and $b$ be so that as we approach $x = 1$ from either side, we get the same value of $4$?

The key insight is that a limit exists at a point only when the left-hand and right-hand limits agree. Since the function has different expressions on either side of $x = 1$, we need to check both directions separately.

Finding the constraints

1. Left-hand limit as $x \to 1^-$

   When $x < 1$, the function is given by $f(x) = a + bx$. As $x$ approaches $1$ from the left, we substitute directly into this expression:

2. Right-hand limit as $x \to 1^+$

   When $x > 1$, the function is $f(x) = b - ax$. Approaching $1$ from the right:

3. Setting up the system

   For the overall limit to exist and equal $f(1) = 4$, we need:

This gives us two equations:

4. Solving the system

   Add the two equations:

$2b = 8$ …