91Ó°ÊÓ

First, check if substituting the limit value, \(y = 2\), into the expression results in an indeterminate form (such as \(\frac{0}{0}\), or \(\frac{\infty}{\infty}\)): $$\frac{(2)^{2} + (2) - 6}{\sqrt{8 - (2)^{2}} - (2)} = \frac{0}{0}$$ Since the expression's current form is indeterminate, we need to find a way to simplify it.

Let's try factoring the numerator to get a better understanding of the expression: $$\frac{y^{2} + y - 6}{\sqrt{8 - y^{2}} - y} = \frac{(y + 3)(y - 2)}{\sqrt{8 - y^{2}} - y}$$ Now we should find a way to cancel out the \((y - 2)\) term in the numerator to make the expression more manageable. Here, we'll multiply both the numerator and denominator by the conjugate of the denominator: $$\frac{(y + 3)(y - 2)}{\sqrt{8 - y^{2}} - y} \cdot \frac{\sqrt{8 - y^{2}} + y}{\sqrt{8 - y^{2}} + y} = \frac{(y + 3)(y - 2)(\sqrt{8 - y^{2}} + y)}{(8 - y^{2}) - y^2}$$ Simplify the denominator: $$\frac{(y + 3)(y - 2)(\sqrt{8 - y^{2}} + y)}{-y^2 + 8} = \frac{(y + 3)(\sqrt{8 - y^{2}} + y)}{-1}$$ Now, we see that the \((y - 2)\) term from the numerator has been effectively canceled out.

Now that we have a simplified expression, substitute the limit value \(y = 2\) again to find the value of the limit: $$\lim_{y \rightarrow 2} \frac{(2 + 3)(\sqrt{8 - (2)^{2}} + 2)}{-1} = \frac{5 \cdot (2 + 2)}{-1} = \frac{20}{-1} = -20$$ So, we find that the limit is: $$\lim_{y \rightarrow 2} \frac{y^{2} + y - 6}{\sqrt{8 - y^{2}} - y} = -20$$