Question

limits quiz
a
\\(\lim\limits_{x \to 2} (8 - 3x + 12x^2)\\)
b
\\(\lim\limits_{x \to 8} \frac{2x^2 - 17x + 8}{8 - x}\\)
c
\\(\lim\limits_{t \to -3} \frac{6 + 4t}{t^2 + 1}\\)
d
\\(\lim\limits_{x \to 5} (10 + |x - 5|)\\)
e
\\(\lim\limits_{x \to 5} \frac{x + 8}{x - 5}\\)
f
\\(\lim\limits_{x \to 5} \frac{2x^2 - 12x - 10}{2x - 10}\\)
g
\\(\lim\limits_{x \to 3} \frac{x^2 - 6x + 9}{x - 3}\\)
h
\\(\lim\limits_{x \to 2} \frac{x^2 + 8x + 16}{x + 4}\\)
challenge question
i
\\(\lim\limits_{x \to -5} \frac{x^2 - 25}{x^2 + 2x - 15}\\)
j
\\(\lim\limits_{w \to -4} \frac{w^2 - 16}{(w - 2)(w + 3) - 8}\\)

Explanation:

A:

Step1: Substitute $x=2$

$\lim_{x \to 2}(8 - 3x + 12x^2) = 8 - 3(2) + 12(2)^2$

Step2: Calculate the value

$= 8 - 6 + 12(4) = 2 + 48 = 50$

B:

Step1: Factor numerator

$2x^2 -17x +8 = (2x-1)(x-8)$
$\lim_{x \to 8}\frac{(2x-1)(x-8)}{8-x} = \lim_{x \to 8}\frac{(2x-1)(x-8)}{-(x-8)}$

Step2: Cancel common terms

$\lim_{x \to 8} -(2x-1)$

Step3: Substitute $x=8$

$= -(2(8)-1) = -(16-1) = -15$

C:

Step1: Substitute $t \to -\infty$

For large $|t|$, $\frac{6+4t}{t^2+1} \approx \frac{4t}{t^2} = \frac{4}{t}$

Step2: Evaluate the limit

$\lim_{t \to -\infty}\frac{4}{t} = 0$

D:

Step1: Substitute $x=5$

$\lim_{x \to 5}(10 + |x-5|) = 10 + |5-5|$

Step2: Calculate absolute value

$= 10 + 0 = 10$

E:

Step1: Check left/right limits

Right limit ($x \to 5^+$): $\lim_{x \to 5^+}\frac{x+5}{x-5} = +\infty$
Left limit ($x \to 5^-$): $\lim_{x \to 5^-}\frac{x+5}{x-5} = -\infty$

Step2: Conclude limit status

Left $
eq$ right limit, so limit does not exist.

F:

Step1: Factor numerator/denominator

$2x^2-13x-10=(2x+2)(x-5)$, $2x-10=2(x-5)$
$\lim_{x \to 5}\frac{(2x+2)(x-5)}{2(x-5)}$

Step2: Cancel common terms

$\lim_{x \to 5}\frac{2x+2}{2} = \lim_{x \to 5}(x+1)$

Step3: Substitute $x=5$

$= 5+1=6$

G:

Step1: Factor numerator

$x^2-6x+9=(x-3)^2$
$\lim_{x \to 3}\frac{(x-3)^2}{x-3} = \lim_{x \to 3}(x-3)$

Step2: Substitute $x=3$

$= 3-3=0$

H:

Step1: Factor numerator/denominator

$x^2+8x+16=(x+4)^2$, $x+4$ is the denominator
$\lim_{x \to -2}\frac{(x+4)^2}{x+4} = \lim_{x \to -2}(x+4)$

Step2: Substitute $x=-2$

$= -2+4=2$

I:

Step1: Factor numerator/denominator

$x^2-25=(x-5)(x+5)$, $x^2+2x-15=(x+5)(x-3)$
$\lim_{x \to -5}\frac{(x-5)(x+5)}{(x+5)(x-3)}$

Step2: Cancel common terms

$\lim_{x \to -5}\frac{x-5}{x-3}$

Step3: Substitute $x=-5$

$= \frac{-5-5}{-5-3} = \frac{-10}{-8} = \frac{5}{4}$

J:

Step1: Simplify denominator

$(w-2)(w+3)-6 = w^2+w-6-6 = w^2+w-12=(w+4)(w-3)$
Numerator: $w^2-16=(w-4)(w+4)$
$\lim_{w \to -4}\frac{(w-4)(w+4)}{(w+4)(w-3)}$

Step2: Cancel common terms

$\lim_{w \to -4}\frac{w-4}{w-3}$

Step3: Substitute $w=-4$

$= \frac{-4-4}{-4-3} = \frac{-8}{-7} = \frac{8}{7}$

Answer:

A. $\boldsymbol{50}$
B. $\boldsymbol{-15}$
C. $\boldsymbol{0}$
D. $\boldsymbol{10}$
E. $\text{Does not exist}$
F. $\boldsymbol{6}$
G. $\boldsymbol{0}$
H. $\boldsymbol{2}$
I. $\boldsymbol{\frac{5}{4}}$
J. $\boldsymbol{\frac{8}{7}}$