lim y-9 (y−3)/(y2-8y−9)

Question

lim y->9 (y−3)/(y2-8y−9)

ScanSolve AI Tutor · Accepted Answer

To evaluate the limit, first, substitute $y=9$ into the expression to check for an indeterminate form.

\frac{\sqrt{9}-3}{9^2-8(9)-9} = \frac{3-3}{81-72-9} = \frac{0}{0}

Since we have the indeterminate form $\frac{0}{0}$ , we need to simplify the expression.

Factor the denominator.
The denominator is a quadratic expression $y^2 - 8y - 9$ . We look for two numbers that multiply to -9 and add to -8. These numbers are -9 and 1.
$y^2 - 8y - 9 = (y-9)(y+1)$
Rationalize the numerator.
Multiply the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{y}+3$ .
$\lim_{y \to 9} \frac{\sqrt{y}-3}{(y-9)(y+1)} \times \frac{\sqrt{y}+3}{\sqrt{y}+3}$
Simplify the numerator using the difference of squares formula $(a-b)(a+b) = a^2-b^2$ .
$(\sqrt{y}-3)(\sqrt{y}+3) = (\sqrt{y})^2 - 3^2 = y-9$
Substitute the simplified numerator back into the limit expression.
$\lim_{y \to 9} \frac{y-9}{(y-9)(y+1)(\sqrt{y}+3)}$
Cancel out the common factor $(y-9)$ from the numerator and denominator.
Since $y \to 9$ , $y \neq 9$ , so $y-9 \neq 0$ .
$\lim_{y \to 9} \frac{1}{(y+1)(\sqrt{y}+3)}$
Substitute $y=9$ into the simplified expression.
$\frac{1}{(9+1)(\sqrt{9}+3)} = \frac{1}{(10)(3+3)} = \frac{1}{(10)(6)} = \frac{1}{60}$
The limit is:
$\frac{1}{60}$

lim y->9 (y-3)/(y2-8y-9)