Math  /  Calculus

QuestionDetermining Limits Determine the following limits: a) limx2(x24)\quad \lim _{x \rightarrow 2}\left(x^{2}-4\right) j) limx2x2x23x+2\quad \lim _{x \rightarrow 2} \frac{x-2}{x^{2}-3 x+2} s) limx2x3+3x2+2xx2x6\lim _{x \rightarrow-2} \frac{x^{3}+3 x^{2}+2 x}{x^{2}-x-6} b) limx0x34x2x2+3x\lim _{x \rightarrow 0} \frac{x^{3}-4 x}{2 x^{2}+3 x} k) limx03x+2x1x+4x1\lim _{x \rightarrow 0} \frac{3 x+2 x^{-1}}{x+4 x^{-1}} t) limx1x22x+1x3x\lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{x^{3}-x}

Studdy Solution

STEP 1

1. We are given several limit problems to solve.
2. We will use algebraic simplification and limit properties to find the limits.
3. We assume the functions are defined and continuous near the points of interest unless otherwise stated.

STEP 2

1. Solve limit (a) using direct substitution.
2. Solve limit (j) using algebraic simplification.
3. Solve limit (s) using algebraic simplification.
4. Solve limit (b) using algebraic simplification.
5. Solve limit (k) using algebraic simplification.
6. Solve limit (t) using algebraic simplification.

STEP 3

For limit (a), use direct substitution:
limx2(x24)=(2)24 \lim_{x \rightarrow 2} (x^2 - 4) = (2)^2 - 4

STEP 4

Calculate the expression:
44=0 4 - 4 = 0
So,
limx2(x24)=0 \lim_{x \rightarrow 2} (x^2 - 4) = 0

STEP 5

For limit (j), factor the numerator and denominator:
limx2x2x23x+2 \lim_{x \rightarrow 2} \frac{x-2}{x^2 - 3x + 2}
Factor the denominator:
x23x+2=(x1)(x2) x^2 - 3x + 2 = (x-1)(x-2)

STEP 6

Cancel common factors:
limx2x2(x1)(x2)=limx21x1 \lim_{x \rightarrow 2} \frac{x-2}{(x-1)(x-2)} = \lim_{x \rightarrow 2} \frac{1}{x-1}

STEP 7

Substitute x=2 x = 2 :
121=1 \frac{1}{2-1} = 1
So,
limx2x2x23x+2=1 \lim_{x \rightarrow 2} \frac{x-2}{x^2 - 3x + 2} = 1

STEP 8

For limit (s), factor both numerator and denominator:
limx2x3+3x2+2xx2x6 \lim_{x \rightarrow -2} \frac{x^3 + 3x^2 + 2x}{x^2 - x - 6}
Factor the numerator:
x(x2+3x+2)=x(x+1)(x+2) x(x^2 + 3x + 2) = x(x+1)(x+2)
Factor the denominator:
x2x6=(x3)(x+2) x^2 - x - 6 = (x-3)(x+2)

STEP 9

Cancel common factors:
limx2x(x+1)(x+2)(x3)(x+2)=limx2x(x+1)x3 \lim_{x \rightarrow -2} \frac{x(x+1)(x+2)}{(x-3)(x+2)} = \lim_{x \rightarrow -2} \frac{x(x+1)}{x-3}

STEP 10

Substitute x=2 x = -2 :
2(2+1)23=2(1)5=25 \frac{-2(-2+1)}{-2-3} = \frac{-2(-1)}{-5} = \frac{2}{5}
So,
limx2x3+3x2+2xx2x6=25 \lim_{x \rightarrow -2} \frac{x^3 + 3x^2 + 2x}{x^2 - x - 6} = \frac{2}{5}

STEP 11

For limit (b), factor both numerator and denominator:
limx0x34x2x2+3x \lim_{x \rightarrow 0} \frac{x^3 - 4x}{2x^2 + 3x}
Factor the numerator:
x(x24)=x(x2)(x+2) x(x^2 - 4) = x(x-2)(x+2)
Factor the denominator:
x(2x+3) x(2x + 3)

STEP 12

Cancel common factors:
limx0x(x2)(x+2)x(2x+3)=limx0(x2)(x+2)2x+3 \lim_{x \rightarrow 0} \frac{x(x-2)(x+2)}{x(2x+3)} = \lim_{x \rightarrow 0} \frac{(x-2)(x+2)}{2x+3}

STEP 13

Substitute x=0 x = 0 :
(02)(0+2)2(0)+3=(2)(2)3=43 \frac{(0-2)(0+2)}{2(0)+3} = \frac{(-2)(2)}{3} = \frac{-4}{3}
So,
limx0x34x2x2+3x=43 \lim_{x \rightarrow 0} \frac{x^3 - 4x}{2x^2 + 3x} = \frac{-4}{3}

STEP 14

For limit (k), simplify the expression:
limx03x+2x1x+4x1 \lim_{x \rightarrow 0} \frac{3x + 2x^{-1}}{x + 4x^{-1}}
Rewrite using common denominators:
limx03x2+2xx2+4x=limx03x2+2x2+4 \lim_{x \rightarrow 0} \frac{\frac{3x^2 + 2}{x}}{\frac{x^2 + 4}{x}} = \lim_{x \rightarrow 0} \frac{3x^2 + 2}{x^2 + 4}

STEP 15

Substitute x=0 x = 0 :
3(0)2+202+4=24=12 \frac{3(0)^2 + 2}{0^2 + 4} = \frac{2}{4} = \frac{1}{2}
So,
limx03x+2x1x+4x1=12 \lim_{x \rightarrow 0} \frac{3x + 2x^{-1}}{x + 4x^{-1}} = \frac{1}{2}

STEP 16

For limit (t), factor both numerator and denominator:
limx1x22x+1x3x \lim_{x \rightarrow 1} \frac{x^2 - 2x + 1}{x^3 - x}
Factor the numerator:
(x1)2 (x-1)^2
Factor the denominator:
x(x21)=x(x1)(x+1) x(x^2 - 1) = x(x-1)(x+1)

STEP 17

Cancel common factors:
limx1(x1)2x(x1)(x+1)=limx1x1x(x+1) \lim_{x \rightarrow 1} \frac{(x-1)^2}{x(x-1)(x+1)} = \lim_{x \rightarrow 1} \frac{x-1}{x(x+1)}

STEP 18

Substitute x=1 x = 1 :
111(1+1)=02=0 \frac{1-1}{1(1+1)} = \frac{0}{2} = 0
So,
limx1x22x+1x3x=0 \lim_{x \rightarrow 1} \frac{x^2 - 2x + 1}{x^3 - x} = 0
The solutions to the limits are:
a) limx2(x24)=0 \lim_{x \rightarrow 2} (x^2 - 4) = 0
j) limx2x2x23x+2=1 \lim_{x \rightarrow 2} \frac{x-2}{x^2 - 3x + 2} = 1
s) limx2x3+3x2+2xx2x6=25 \lim_{x \rightarrow -2} \frac{x^3 + 3x^2 + 2x}{x^2 - x - 6} = \frac{2}{5}
b) limx0x34x2x2+3x=43 \lim_{x \rightarrow 0} \frac{x^3 - 4x}{2x^2 + 3x} = \frac{-4}{3}
k) limx03x+2x1x+4x1=12 \lim_{x \rightarrow 0} \frac{3x + 2x^{-1}}{x + 4x^{-1}} = \frac{1}{2}
t) limx1x22x+1x3x=0 \lim_{x \rightarrow 1} \frac{x^2 - 2x + 1}{x^3 - x} = 0

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