\( \lim _ { x \rightarrow 4 } \frac { x + 1 } { x - 2 \sqrt { a x + b } } = + \infty \quad b = ? \)

To find the value of \( b \) such that \[ \lim _ { x \rightarrow 4 } \frac { x + 1 } { x - 2 \sqrt { a x + b } } = + \infty, \] we need the denominator to approach zero as \( x \) approaches 4, while the numerator approaches a non-zero value. First, let's analyze the numerator: As \( x \to 4 \): \[ x + 1 \to 5. \] Thus, for the limit to equal \( +\infty \), the denominator must approach zero from the positive side. Therefore, we need: \[ x - 2 \sqrt{a x + b} \to 0 \text{ as } x \to 4. \] Setting \( x = 4 \): \[ 4 - 2 \sqrt{4a + b} = 0 \quad \Rightarrow \quad 2 \sqrt{4a + b} = 4 \quad \Rightarrow \quad \sqrt{4a + b} = 2. \] Squaring both sides gives: \[ 4a + b = 4. \] Next, we isolate \( b \): \[ b = 4 - 4a. \] So the value of \( b \) can be expressed in terms of \( a \). If we want a specific value for \( b \), we can choose a value of \( a \) to find it. For instance, if we let \( a = 1 \): \[ b = 4 - 4(1) = 0. \] Therefore, one possible value for \( b \) is 0. However, \( b \) could take on other values depending on \( a \), holding to the relationship \( b = 4 - 4a \).