Math / Calculus

Question3.1.2
In Problems 37-54, use the limit laws to evaluate each limit.
37. $\lim _{x \rightarrow-1}\left(x^{3}+7 x-1\right)$
39. $\lim _{x \rightarrow-5}\left(4+2 x^{2}\right)$
38. $\lim _{x \rightarrow 2}\left(3 x^{4}-2 x+1\right)$
41. $\lim _{x \rightarrow 3}\left(2 x^{2}-\frac{1}{x}\right)$
40. $\lim _{x \rightarrow 2}\left(8 x^{3}-2 x+4\right)$
43. $\lim _{x \rightarrow-3} \frac{x^{3}-20}{x+1}$
42. $\lim _{x \rightarrow-2}\left(\frac{x^{2}}{2}-\frac{2}{x^{2}}\right)$
45. $\lim _{x \rightarrow 3} \frac{3 x^{2}+1}{2 x-3}$
44. $\lim _{x \rightarrow 1} \frac{x^{3}-1}{x+2}$
47. $\lim _{x \rightarrow 1} \frac{1-x^{2}}{1-x}$
46. $\lim _{x \rightarrow-2} \frac{1+x}{1-x}$
49. $\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3}$
48. $\lim _{u \rightarrow 3} \frac{9-u^{2}}{3-u}$
51. $\lim _{x \rightarrow 2} \frac{2-x}{x^{2}-4}$
50. $\lim _{x \rightarrow 1} \frac{(x-1)^{2}}{x^{2}-1}$
53. $\lim _{x \rightarrow-2} \frac{2 x^{2}+3 x-2}{x+2}$
52. $\lim _{x \rightarrow-4} \frac{x+4}{16-x^{2}}$
54. $\lim _{x \rightarrow 1 / 2} \frac{1-x-2 x^{2}}{1-2 x}$

Studdy Solution

STEP 1

1. We are given several limits to evaluate using limit laws.
2. We assume that all functions are continuous at the points of evaluation unless otherwise specified.
3. We will apply limit laws such as the sum, difference, product, quotient, and power laws as needed.

STEP 2

1. Evaluate polynomial limits directly by substitution.
2. Simplify rational expressions if necessary and evaluate limits.
3. Apply limit laws to evaluate limits involving fractions.
Let's solve each problem one by one.

STEP 3

Evaluate $\lim _{x \rightarrow-1}\left(x^{3}+7 x-1\right)$ :
Direct substitution is possible since the function is a polynomial:
$\lim_{x \to -1} (x^3 + 7x - 1) = (-1)^3 + 7(-1) - 1$

STEP 4

Simplify the expression:
$= -1 - 7 - 1$ $= -9$

STEP 5

Evaluate $\lim _{x \rightarrow-5}\left(4+2 x^{2}\right)$ :
Direct substitution is possible since the function is a polynomial:
$\lim_{x \to -5} (4 + 2x^2) = 4 + 2(-5)^2$

STEP 6

Simplify the expression:
$= 4 + 2(25)$ $= 4 + 50$ $= 54$

STEP 7

Evaluate $\lim _{x \rightarrow 2}\left(3 x^{4}-2 x+1\right)$ :
Direct substitution is possible since the function is a polynomial:
$\lim_{x \to 2} (3x^4 - 2x + 1) = 3(2)^4 - 2(2) + 1$

STEP 8

Simplify the expression:
$= 3(16) - 4 + 1$ $= 48 - 4 + 1$ $= 45$

STEP 9

Evaluate $\lim _{x \rightarrow 3}\left(2 x^{2}-\frac{1}{x}\right)$ :
Direct substitution is possible since the function is a polynomial and rational function:
$\lim_{x \to 3} \left(2x^2 - \frac{1}{x}\right) = 2(3)^2 - \frac{1}{3}$

STEP 10

Simplify the expression:
$= 2(9) - \frac{1}{3}$ $= 18 - \frac{1}{3}$ $= \frac{54}{3} - \frac{1}{3}$ $= \frac{53}{3}$

STEP 11

Evaluate $\lim _{x \rightarrow 2}\left(8 x^{3}-2 x+4\right)$ :
Direct substitution is possible since the function is a polynomial:
$\lim_{x \to 2} (8x^3 - 2x + 4) = 8(2)^3 - 2(2) + 4$

STEP 12

Simplify the expression:
$= 8(8) - 4 + 4$ $= 64 - 4 + 4$ $= 64$

STEP 13

Evaluate $\lim _{x \rightarrow-3} \frac{x^{3}-20}{x+1}$ :
Direct substitution results in an indeterminate form, so we need to simplify:
$\lim_{x \to -3} \frac{x^3 - 20}{x + 1}$

STEP 14

Since direct substitution doesn't work, check for factorization or other simplification techniques. Since $x^3 - 20$ does not factor nicely, use direct substitution to check if the limit exists:
$\lim_{x \to -3} \frac{(-3)^3 - 20}{-3 + 1} = \frac{-27 - 20}{-2}$

STEP 15

Simplify the expression:
$= \frac{-47}{-2}$ $= \frac{47}{2}$

STEP 16

Evaluate $\lim _{x \rightarrow-2}\left(\frac{x^{2}}{2}-\frac{2}{x^{2}}\right)$ :
Direct substitution is possible since the function is a polynomial and rational function:
$\lim_{x \to -2} \left(\frac{x^2}{2} - \frac{2}{x^2}\right) = \frac{(-2)^2}{2} - \frac{2}{(-2)^2}$

STEP 17

Simplify the expression:
$= \frac{4}{2} - \frac{2}{4}$ $= 2 - \frac{1}{2}$ $= \frac{4}{2} - \frac{1}{2}$ $= \frac{3}{2}$

STEP 18

Evaluate $\lim _{x \rightarrow 3} \frac{3 x^{2}+1}{2 x-3}$ :
Direct substitution results in a non-zero denominator, so evaluate directly:
$\lim_{x \to 3} \frac{3x^2 + 1}{2x - 3} = \frac{3(3)^2 + 1}{2(3) - 3}$

STEP 19

Simplify the expression:
$= \frac{27 + 1}{6 - 3}$ $= \frac{28}{3}$

STEP 20

Evaluate $\lim _{x \rightarrow 1} \frac{x^{3}-1}{x+2}$ :
Direct substitution is possible since the function is a polynomial and rational function:
$\lim_{x \to 1} \frac{x^3 - 1}{x + 2} = \frac{1^3 - 1}{1 + 2}$

STEP 21

Simplify the expression:
$= \frac{0}{3}$ $= 0$

STEP 22

Evaluate $\lim _{x \rightarrow 1} \frac{1-x^{2}}{1-x}$ :
Direct substitution results in an indeterminate form, so factor the numerator:
$\lim_{x \to 1} \frac{1 - x^2}{1 - x} = \lim_{x \to 1} \frac{(1 - x)(1 + x)}{1 - x}$

STEP 23

Cancel the common factor:
$= \lim_{x \to 1} (1 + x)$

STEP 24

Substitute $x = 1$ :
$= 1 + 1$ $= 2$

STEP 25

Evaluate $\lim _{x \rightarrow-2} \frac{1+x}{1-x}$ :
Direct substitution is possible since the function is a polynomial and rational function:
$\lim_{x \to -2} \frac{1 + x}{1 - x} = \frac{1 + (-2)}{1 - (-2)}$

STEP 26

Simplify the expression:
$= \frac{-1}{3}$

STEP 27

Evaluate $\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3}$ :
Direct substitution results in an indeterminate form, so factor the numerator:
$\lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 1)}{x - 3}$

STEP 28

Cancel the common factor:
$= \lim_{x \to 3} (x + 1)$

STEP 29

Substitute $x = 3$ :
$= 3 + 1$ $= 4$

STEP 30

Evaluate $\lim _{u \rightarrow 3} \frac{9-u^{2}}{3-u}$ :
Direct substitution results in an indeterminate form, so factor the numerator:
$\lim_{u \to 3} \frac{9 - u^2}{3 - u} = \lim_{u \to 3} \frac{(3 - u)(3 + u)}{3 - u}$

STEP 31

Cancel the common factor:
$= \lim_{u \to 3} (3 + u)$

STEP 32

Substitute $u = 3$ :
$= 3 + 3$ $= 6$

STEP 33

Evaluate $\lim _{x \rightarrow 2} \frac{2-x}{x^{2}-4}$ :
Direct substitution results in an indeterminate form, so factor the denominator:
$\lim_{x \to 2} \frac{2 - x}{x^2 - 4} = \lim_{x \to 2} \frac{2 - x}{(x - 2)(x + 2)}$

STEP 34

Notice that $2 - x = -(x - 2)$ , so rewrite the expression:
$= \lim_{x \to 2} \frac{-(x - 2)}{(x - 2)(x + 2)}$

STEP 35

Cancel the common factor:
$= \lim_{x \to 2} \frac{-1}{x + 2}$

STEP 36

Substitute $x = 2$ :
$= \frac{-1}{2 + 2}$ $= \frac{-1}{4}$

STEP 37

Evaluate $\lim _{x \rightarrow 1} \frac{(x-1)^{2}}{x^{2}-1}$ :
Direct substitution results in an indeterminate form, so factor the denominator:
$\lim_{x \to 1} \frac{(x - 1)^2}{x^2 - 1} = \lim_{x \to 1} \frac{(x - 1)^2}{(x - 1)(x + 1)}$

STEP 38

Cancel the common factor:
$= \lim_{x \to 1} \frac{x - 1}{x + 1}$

STEP 39

Substitute $x = 1$ :
$= \frac{1 - 1}{1 + 1}$ $= \frac{0}{2}$ $= 0$

STEP 40

Evaluate $\lim _{x \rightarrow-4} \frac{x+4}{16-x^{2}}$ :
Direct substitution results in an indeterminate form, so factor the denominator:
$\lim_{x \to -4} \frac{x + 4}{16 - x^2} = \lim_{x \to -4} \frac{x + 4}{(4 - x)(4 + x)}$

STEP 41

Notice that $16 - x^2 = (4 - x)(4 + x)$ , so rewrite the expression:
$= \lim_{x \to -4} \frac{1}{4 - x}$

STEP 42

Substitute $x = -4$ :
$= \frac{1}{4 + 4}$ $= \frac{1}{8}$

STEP 43

Evaluate $\lim _{x \rightarrow 1 / 2} \frac{1-x-2 x^{2}}{1-2 x}$ :
Direct substitution is possible since the function is a polynomial and rational function:
$\lim_{x \to \frac{1}{2}} \frac{1 - x - 2x^2}{1 - 2x} = \frac{1 - \frac{1}{2} - 2\left(\frac{1}{2}\right)^2}{1 - 2\left(\frac{1}{2}\right)}$

STEP 44

Simplify the expression:
$= \frac{1 - \frac{1}{2} - \frac{1}{2}}{1 - 1}$
Since the denominator is zero, the limit does not exist.

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Studdy Solution

STEP 1

1. We are given several limits to evaluate using limit laws.2. We assume that all functions are continuous at the points of evaluation unless otherwise specified.3. We will apply limit laws such as the sum, difference, product, quotient, and power laws as needed.

STEP 2

1. Evaluate polynomial limits directly by substitution.2. Simplify rational expressions if necessary and evaluate limits.3. Apply limit laws to evaluate limits involving fractions.Let's solve each problem one by one.

STEP 3

STEP 4

Simplify the expression:=−1−7−1 = -1 - 7 - 1 =−1−7−1 =−9 = -9 =−9

STEP 5

Evaluate lim⁡x→−5(4+2x2)\lim _{x \rightarrow-5}\left(4+2 x^{2}\right)limx→−5​(4+2x2):Direct substitution is possible since the function is a polynomial:lim⁡x→−5(4+2x2)=4+2(−5)2 \lim_{x \to -5} (4 + 2x^2) = 4 + 2(-5)^2 x→−5lim​(4+2x2)=4+2(−5)2

STEP 6

Simplify the expression:=4+2(25) = 4 + 2(25) =4+2(25) =4+50 = 4 + 50 =4+50 =54 = 54 =54

STEP 7

STEP 8

Simplify the expression:=3(16)−4+1 = 3(16) - 4 + 1 =3(16)−4+1 =48−4+1 = 48 - 4 + 1 =48−4+1 =45 = 45 =45

STEP 9

STEP 10

Simplify the expression:=2(9)−13 = 2(9) - \frac{1}{3} =2(9)−31​ =18−13 = 18 - \frac{1}{3} =18−31​ =543−13 = \frac{54}{3} - \frac{1}{3} =354​−31​ =533 = \frac{53}{3} =353​

STEP 11

STEP 12

Simplify the expression:=8(8)−4+4 = 8(8) - 4 + 4 =8(8)−4+4 =64−4+4 = 64 - 4 + 4 =64−4+4 =64 = 64 =64

STEP 13

Evaluate lim⁡x→−3x3−20x+1\lim _{x \rightarrow-3} \frac{x^{3}-20}{x+1}limx→−3​x+1x3−20​:Direct substitution results in an indeterminate form, so we need to simplify:lim⁡x→−3x3−20x+1 \lim_{x \to -3} \frac{x^3 - 20}{x + 1} x→−3lim​x+1x3−20​

STEP 14

STEP 15

Simplify the expression:=−47−2 = \frac{-47}{-2} =−2−47​ =472 = \frac{47}{2} =247​

STEP 16

STEP 17

Simplify the expression:=42−24 = \frac{4}{2} - \frac{2}{4} =24​−42​ =2−12 = 2 - \frac{1}{2} =2−21​ =42−12 = \frac{4}{2} - \frac{1}{2} =24​−21​ =32 = \frac{3}{2} =23​

STEP 18

STEP 19

Simplify the expression:=27+16−3 = \frac{27 + 1}{6 - 3} =6−327+1​ =283 = \frac{28}{3} =328​

STEP 20

STEP 21

Simplify the expression:=03 = \frac{0}{3} =30​ =0 = 0 =0

STEP 22

STEP 23

Cancel the common factor:=lim⁡x→1(1+x) = \lim_{x \to 1} (1 + x) =x→1lim​(1+x)

STEP 24

Substitute x=1x = 1x=1:=1+1 = 1 + 1 =1+1 =2 = 2 =2

STEP 25

STEP 26

Simplify the expression:=−13 = \frac{-1}{3} =3−1​

STEP 27

STEP 28

Cancel the common factor:=lim⁡x→3(x+1) = \lim_{x \to 3} (x + 1) =x→3lim​(x+1)

STEP 29

Substitute x=3x = 3x=3:=3+1 = 3 + 1 =3+1 =4 = 4 =4

STEP 30

STEP 31

Cancel the common factor:=lim⁡u→3(3+u) = \lim_{u \to 3} (3 + u) =u→3lim​(3+u)

STEP 32

Substitute u=3u = 3u=3:=3+3 = 3 + 3 =3+3 =6 = 6 =6

STEP 33

STEP 34

Notice that 2−x=−(x−2)2 - x = -(x - 2)2−x=−(x−2), so rewrite the expression:=lim⁡x→2−(x−2)(x−2)(x+2) = \lim_{x \to 2} \frac{-(x - 2)}{(x - 2)(x + 2)} =x→2lim​(x−2)(x+2)−(x−2)​

STEP 35

Cancel the common factor:=lim⁡x→2−1x+2 = \lim_{x \to 2} \frac{-1}{x + 2} =x→2lim​x+2−1​

STEP 36

Substitute x=2x = 2x=2:=−12+2 = \frac{-1}{2 + 2} =2+2−1​ =−14 = \frac{-1}{4} =4−1​

STEP 37

STEP 38

Cancel the common factor:=lim⁡x→1x−1x+1 = \lim_{x \to 1} \frac{x - 1}{x + 1} =x→1lim​x+1x−1​

STEP 39

Substitute x=1x = 1x=1:=1−11+1 = \frac{1 - 1}{1 + 1} =1+11−1​ =02 = \frac{0}{2} =20​ =0 = 0 =0

STEP 40

STEP 41

Notice that 16−x2=(4−x)(4+x)16 - x^2 = (4 - x)(4 + x)16−x2=(4−x)(4+x), so rewrite the expression:=lim⁡x→−414−x = \lim_{x \to -4} \frac{1}{4 - x} =x→−4lim​4−x1​

STEP 42

Substitute x=−4x = -4x=−4:=14+4 = \frac{1}{4 + 4} =4+41​ =18 = \frac{1}{8} =81​

STEP 43

STEP 44

Simplify the expression:=1−12−121−1 = \frac{1 - \frac{1}{2} - \frac{1}{2}}{1 - 1} =1−11−21​−21​​Since the denominator is zero, the limit does not exist.

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1. We are given several limits to evaluate using limit laws.
2. We assume that all functions are continuous at the points of evaluation unless otherwise specified.
3. We will apply limit laws such as the sum, difference, product, quotient, and power laws as needed.

1. Evaluate polynomial limits directly by substitution.
2. Simplify rational expressions if necessary and evaluate limits.
3. Apply limit laws to evaluate limits involving fractions.
Let's solve each problem one by one.

Simplify the expression:
$= -1 - 7 - 1$ $= -9$

Evaluate $\lim _{x \rightarrow-5}\left(4+2 x^{2}\right)$ :
Direct substitution is possible since the function is a polynomial:
$\lim_{x \to -5} (4 + 2x^2) = 4 + 2(-5)^2$

Simplify the expression:
$= 4 + 2(25)$ $= 4 + 50$ $= 54$

Simplify the expression:
$= 3(16) - 4 + 1$ $= 48 - 4 + 1$ $= 45$

Simplify the expression:
$= 2(9) - \frac{1}{3}$ $= 18 - \frac{1}{3}$ $= \frac{54}{3} - \frac{1}{3}$ $= \frac{53}{3}$

Simplify the expression:
$= 8(8) - 4 + 4$ $= 64 - 4 + 4$ $= 64$

Evaluate $\lim _{x \rightarrow-3} \frac{x^{3}-20}{x+1}$ :
Direct substitution results in an indeterminate form, so we need to simplify:
$\lim_{x \to -3} \frac{x^3 - 20}{x + 1}$

Simplify the expression:
$= \frac{-47}{-2}$ $= \frac{47}{2}$

Simplify the expression:
$= \frac{4}{2} - \frac{2}{4}$ $= 2 - \frac{1}{2}$ $= \frac{4}{2} - \frac{1}{2}$ $= \frac{3}{2}$

Simplify the expression:
$= \frac{27 + 1}{6 - 3}$ $= \frac{28}{3}$

Simplify the expression:
$= \frac{0}{3}$ $= 0$

Cancel the common factor:
$= \lim_{x \to 1} (1 + x)$

Substitute $x = 1$ :
$= 1 + 1$ $= 2$

Simplify the expression:
$= \frac{-1}{3}$

Cancel the common factor:
$= \lim_{x \to 3} (x + 1)$

Substitute $x = 3$ :
$= 3 + 1$ $= 4$

Cancel the common factor:
$= \lim_{u \to 3} (3 + u)$

Substitute $u = 3$ :
$= 3 + 3$ $= 6$

Notice that $2 - x = -(x - 2)$ , so rewrite the expression:
$= \lim_{x \to 2} \frac{-(x - 2)}{(x - 2)(x + 2)}$

Cancel the common factor:
$= \lim_{x \to 2} \frac{-1}{x + 2}$

Substitute $x = 2$ :
$= \frac{-1}{2 + 2}$ $= \frac{-1}{4}$

Cancel the common factor:
$= \lim_{x \to 1} \frac{x - 1}{x + 1}$

Substitute $x = 1$ :
$= \frac{1 - 1}{1 + 1}$ $= \frac{0}{2}$ $= 0$

Notice that $16 - x^2 = (4 - x)(4 + x)$ , so rewrite the expression:
$= \lim_{x \to -4} \frac{1}{4 - x}$

Substitute $x = -4$ :
$= \frac{1}{4 + 4}$ $= \frac{1}{8}$

Simplify the expression:
$= \frac{1 - \frac{1}{2} - \frac{1}{2}}{1 - 1}$
Since the denominator is zero, the limit does not exist.