Question

1. evaluate each limit, if it exists.

a) $lim_{x \to 2} \frac{(2+x)^2 - 16}{x - 2}$
b) $lim_{x \to 6} \frac{(3-x)^2 - 9}{x - 6}$
c) $lim_{x \to 2} \frac{49 - (5+x)^2}{x - 2}$
d) $lim_{x \to 3} \frac{\frac{1}{3} - \frac{1}{x}}{x - 3}$
e) $lim_{x \to -2} \frac{x^4 - 16}{x + 2}$
f) $lim_{x \to 1} \frac{x^2 - 1}{x^3 - x^2 - 3x + 3}$

Explanation:

Part a)

Step1: Expand the numerator

$(2+x)^2 - 16 = x^2 +4x +4 -16 = x^2 +4x -12$

Step2: Factor the numerator

$x^2 +4x -12 = (x+6)(x-2)$

Step3: Cancel common factors

$\lim_{x \to 2} \frac{(x+6)(x-2)}{x-2} = \lim_{x \to 2} (x+6)$

Step4: Substitute $x=2$

$2 + 6 = 8$

Part b)

Step1: Expand the numerator

$(3-x)^2 -9 = 9 -6x +x^2 -9 = x^2 -6x$

Step2: Factor the numerator

$x^2 -6x = x(x-6)$

Step3: Cancel common factors

$\lim_{x \to 6} \frac{x(x-6)}{x-6} = \lim_{x \to 6} x$

Step4: Substitute $x=6$

$6$

Part c)

Step1: Expand the numerator

$49-(5+x)^2 = 49 - (x^2 +10x +25) = -x^2 -10x +24$

Step2: Factor the numerator

$-x^2 -10x +24 = -(x^2 +10x -24) = -(x+12)(x-2)$

Step3: Cancel common factors

$\lim_{x \to 2} \frac{-(x+12)(x-2)}{x-2} = \lim_{x \to 2} -(x+12)$

Step4: Substitute $x=2$

$-(2+12) = -14$

Part d)

Step1: Simplify the numerator

$\frac{1}{3} - \frac{1}{x} = \frac{x-3}{3x}$

Step2: Rewrite the limit

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

Step3: Cancel common factors

$\lim_{x \to 3} \frac{1}{3x}$

Step4: Substitute $x=3$

$\frac{1}{3 \times 3} = \frac{1}{9}$

Part e)

Step1: Factor the numerator

$x^4 -16 = (x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4)$

Step2: Cancel common factors

$\lim_{x \to -2} \frac{(x-2)(x+2)(x^2+4)}{x+2} = \lim_{x \to -2} (x-2)(x^2+4)$

Step3: Substitute $x=-2$

$(-2-2)((-2)^2+4) = (-4)(4+4) = (-4)(8) = -32$

Part f)

Step1: Factor numerator and denominator

Numerator: $x^2-1=(x-1)(x+1)$
Denominator: $x^3 -x^2 -3x +3 = x^2(x-1)-3(x-1)=(x^2-3)(x-1)$

Step2: Cancel common factors

$\lim_{x \to 1} \frac{(x-1)(x+1)}{(x^2-3)(x-1)} = \lim_{x \to 1} \frac{x+1}{x^2-3}$

Step3: Substitute $x=1$

$\frac{1+1}{1^2-3} = \frac{2}{-2} = -1$

Answer:

a) $8$
b) $6$
c) $-14$
d) $\frac{1}{9}$
e) $-32$
f) $-1$