Discussion
Let \(f(x) = \begin{cases} ax + 3 & x \lt 1 \\ \frac{4x}{a} & x \ge 1 \end{cases}\). If \(\lim_{x \to 1} f(x)\) exists, then the possible values of \(a\) are
Answer: B
The possible values are 1 and -4.
The limit at x = 1 exists when the left and right limits are equal.
\[ \begin{aligned} a(1) + 3 &= \frac{4(1)}{a} \\ a^{2} + 3a - 4 &= 0 \\ (a - 1)(a + 4) &= 0 \end{aligned} \]
So a = 1 or a = -4. Check: for a = -4 both sides equal -1, and for a = 1 both equal 4. The other pairs are not roots of this quadratic. For a limit to exist the function need not be defined at x = 1 itself; only the two one-sided limits must agree.
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