Question

1. (11pts) evaluate each limit. show all work.

(a) \\(\lim\limits_{x\to 4} \frac{x^2 + x - 20}{-3x + 12}\\)

(b) \\(\lim\limits_{x\to 9} \frac{x - 9}{\sqrt{x} - 3}\\)

(c) \\(\lim\limits_{h\to 0} \frac{(9 + h)^{-1} - 9^{-1}}{h}\\)

Explanation:

(a)

Step1: Factor numerator and denominator

Numerator: $x^2+x-20=(x+5)(x-4)$
Denominator: $-3x+12=-3(x-4)$

Step2: Cancel common factor

$\lim_{x \to 4} \frac{(x+5)(x-4)}{-3(x-4)} = \lim_{x \to 4} \frac{x+5}{-3}$

Step3: Substitute $x=4$

$\frac{4+5}{-3} = \frac{9}{-3}$

(b)

Step1: Rationalize the denominator

Multiply numerator and denominator by $\sqrt{x}+3$:
$\lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}$

Step2: Simplify denominator

$(\sqrt{x}-3)(\sqrt{x}+3)=x-9$

Step3: Cancel common factor

$\lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{x-9} = \lim_{x \to 9} (\sqrt{x}+3)$

Step4: Substitute $x=9$

$\sqrt{9}+3=3+3$

(c)

Step1: Rewrite negative exponents

$(9+h)^{-1}=\frac{1}{9+h}$, $9^{-1}=\frac{1}{9}$

Step2: Combine fractions in numerator

$\frac{1}{9+h}-\frac{1}{9} = \frac{9-(9+h)}{9(9+h)} = \frac{-h}{9(9+h)}$

Step3: Simplify the limit

$\lim_{h \to 0} \frac{\frac{-h}{9(9+h)}}{h} = \lim_{h \to 0} \frac{-1}{9(9+h)}$

Step4: Substitute $h=0$

$\frac{-1}{9(9+0)}$

Answer:

(a) $\boldsymbol{-3}$
(b) $\boldsymbol{6}$
(c) $\boldsymbol{-\frac{1}{81}}$