Solve for b limit as x approaches 4 of (x^2+7x+b)/(x^2-6x+8)=15/2

$lim_{x \to 4} \frac{x^{2} + 7 x + b}{x^{2} - 6 x + 8} = \frac{15}{2}$

Step 1

Since the function approaches $\infty$ from the left and $- \infty$ from the right, the limit does not exist.

$Does not exist = \frac{15}{2}$

Step 2

Move all terms not containing $D$ to the right side of the equation.

Subtract $lim_{x \to 4} \frac{x^{2} + 7 x + b}{x^{2} - 6 x + 8}$ from both sides of the equation.

$0 = \frac{15}{2} - lim_{x \to 4} \frac{x^{2} + 7 x + b}{x^{2} - 6 x + 8}$

Since the function approaches $\infty$ from the left and $- \infty$ from the right, the limit does not exist.

$0 = \frac{15}{2} - Does not exist$

Reorder factors in $\frac{15}{2} - Does not exist$ .

$0 = \frac{15}{2} - lim_{x \to 4} \frac{x^{2} + 7 x + b}{x^{2} - 6 x + 8}$

Step 3

Rewrite the equation as $\frac{15}{2} - lim_{x \to 4} \frac{x^{2} + 7 x + b}{x^{2} - 6 x + 8} = 0$ .

$\frac{15}{2} - lim_{x \to 4} \frac{x^{2} + 7 x + b}{x^{2} - 6 x + 8} = 0$