If $g(x) = \begin{cases} \frac{αx}{|x|}, & \text{if } x < 0 \\ 5, & \text{if } x ≥ 0 \end{cases}$ is continuous at x = 0, then the value of α is

For $g(x)$ to be continuous at $x = 0$:

$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^+} g(x) = g(0)$

Since $x = 0$ satisfies the condition $x ≥ 0$, we use the second part of the function:

$g(0) = 5$

When $x$ approaches $0$ from the right (where $x > 0$), the condition $x ≥ 0$ applies:

$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} 5 = 5$

When $x < 0$, we use $g(x) = \frac{αx}{|x|}$

For negative values of $x$, we have $|x| = -x$

Substituting:

$g(x) = \frac{αx}{|x|}$

$g(x) = \frac{αx}{-x}$

$g(x) = -α$

Therefore:

$\lim_{x \to 0^-} g(x) = -α$

Setting all three values equal:

$-α = 5$

$α = -5$

Therefore, $α = -5$