Question

Let $f(x)=\begin{cases}ax+3 & x<1\\ \frac{4x}{a} & x\ge 1\end{cases}$. If $\lim_{x\rightarrow 1}f(x)$ exists, then the possible values of $a$ are

• A. -1, -4
• B. 1, -4
• C. 4, -4
• D. 4, -1
• E. 1, 4

Hint:

Logic Tip: When evaluating piecewise functions, the existence of a limit at the boundary always generates an algebraic equation by setting the two "pieces" equal to each other at the boundary value.

Answer 1

The Correct Option is B

Concept:
For a limit to exist at a specific point $x=c$, the Left-Hand Limit (LHL) must equal the Right-Hand Limit (RHL) at that point. $$\lim_{x\rightarrow c^{-}} f(x) = \lim_{x\rightarrow c^{+}} f(x)$$
Step 1: Calculate the Left-Hand Limit (LHL).
For values of $x$ approaching 1 from the left ($x<1$), we use the function branch $f(x) = ax + 3$: $$\text{LHL} = \lim_{x\rightarrow 1^{-}} (ax + 3)$$ Substitute $x = 1$: $$\text{LHL} = a(1) + 3 = a + 3$$
Step 2: Calculate the Right-Hand Limit (RHL).
For values of $x$ approaching 1 from the right ($x>1$), we use the function branch $f(x) = \frac{4x}{a}$: $$\text{RHL} = \lim_{x\rightarrow 1^{+}} \left(\frac{4x}{a}\right)$$ Substitute $x = 1$: $$\text{RHL} = \frac{4(1)}{a} = \frac{4}{a}$$
Step 3: Equate LHL and RHL to solve for a.
Since the limit exists, $\text{LHL} = \text{RHL}$: $$a + 3 = \frac{4}{a}$$ Multiply both sides by $a$ (assuming $a \neq 0$): $$a(a + 3) = 4$$ $$a^2 + 3a - 4 = 0$$
Step 4: Factor the quadratic equation.
Find two numbers that multiply to $-4$ and add to $3$. These are $4$ and $-1$: $$(a + 4)(a - 1) = 0$$ This yields two possible values for $a$: $$a = -4 \quad \text{or} \quad a = 1$$