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Evaluate the limits of the following functions:

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Answers

8 parts
a
10
b
55
c
-24
d
2
e

−34-\frac{3}{4}

f
0
g

Does not exist\text{Does not exist}

h

−15-\frac{1}{5}

Step-by-step solution

20 steps

Here are the solutions for the limit problems:

Part a· 2 steps

  1. Substitute x=2x=2 into the function.

    lim⁡x→2(8−3x+2x2)=8−3(2)+2(2)2\lim_{x \to 2} (8 - 3x + 2x^2) = 8 - 3(2) + 2(2)^2

  2. Simplify the expression.

    =8−6+2(4)= 8 - 6 + 2(4) =8−6+8= 8 - 6 + 8 =2+8= 2 + 8 =10= 10

    The limit is 10\boxed{10}.

Part b55

  1. Substitute t=−3t=-3 into the function.

    lim⁡t→−3(1−2t3)=1−2(−3)3\lim_{t \to -3} (1 - 2t^3) = 1 - 2(-3)^3

  2. Simplify the expression.

    =1−2(−27)= 1 - 2(-27) =1+54= 1 + 54 =55= 55

    The limit is 55\boxed{55}.

Part c-24

  1. Substitute y=0y=0 into the function.

    lim⁡y→0(3y4+2y2−3y−24)=3(0)4+2(0)2−3(0)−24\lim_{y \to 0} (3y^4 + 2y^2 - 3y - 24) = 3(0)^4 + 2(0)^2 - 3(0) - 24

  2. Simplify the expression.

    =0+0−0−24= 0 + 0 - 0 - 24 =−24= -24

    The limit is −24\boxed{-24}.

Part d2

  1. Substitute t=−1t=-1 into the function.

    lim⁡t→−13−3t5+2t=3−3(−1)5+2(−1)\lim_{t \to -1} \frac{3 - 3t}{5 + 2t} = \frac{3 - 3(-1)}{5 + 2(-1)}

  2. Simplify the expression.

    =3+35−2= \frac{3 + 3}{5 - 2} =63= \frac{6}{3} =2= 2

    The limit is 2\boxed{2}.

Part e

−34-\frac{3}{4}

  1. Substitute x=−1x=-1 into the function.

    lim⁡x→−1x2−4x2+3=(−1)2−4(−1)2+3\lim_{x \to -1} \frac{x^2 - 4}{x^2 + 3} = \frac{(-1)^2 - 4}{(-1)^2 + 3}

  2. Simplify the expression.

    =1−41+3= \frac{1 - 4}{1 + 3} =−34= \frac{-3}{4}

    The limit is −34\boxed{-\frac{3}{4}}.

Part f0

  1. Substitute y=0y=0 into the function.

    lim⁡y→03y4y+3=3(0)4(0)+3\lim_{y \to 0} \frac{3y}{4y + 3} = \frac{3(0)}{4(0) + 3}

  2. Simplify the expression.

    =00+3= \frac{0}{0 + 3} =03= \frac{0}{3} =0= 0

    The limit is 0\boxed{0}.

Part g

Does not exist\text{Does not exist}

  1. Substitute x=1x=1 into the function.

    lim⁡x→15x+1x−1\lim_{x \to 1} \frac{5x + 1}{x - 1}

    The numerator approaches 5(1)+1=65(1) + 1 = 6.
    The denominator approaches 1−1=01 - 1 = 0.
    Since the numerator is non-zero and the denominator approaches zero, the limit does not exist.

  2. Check left and right-hand limits to determine behavior.

    As x→1+x \to 1^+, x−1>0x-1 > 0, so 5x+1x−1→60+=+∞\frac{5x+1}{x-1} \to \frac{6}{0^+} = +\infty.
    As x→1−x \to 1^-, x−1<0x-1 < 0, so 5x+1x−1→60−=−∞\frac{5x+1}{x-1} \to \frac{6}{0^-} = -\infty.
    Since the left and right-hand limits are not equal, the limit does not exist.
    The limit is Does not exist\boxed{\text{Does not exist}}.

Part h

−15-\frac{1}{5}

  1. Substitute t=−1t=-1 into the function.

    lim⁡t→−110−9t23t2+t−7=10−9(−1)23(−1)2+(−1)−7\lim_{t \to -1} \frac{10 - 9t^2}{3t^2 + t - 7} = \frac{10 - 9(-1)^2}{3(-1)^2 + (-1) - 7}

  2. Simplify the expression.

    =10−9(1)3(1)−1−7= \frac{10 - 9(1)}{3(1) - 1 - 7} =10−93−1−7= \frac{10 - 9}{3 - 1 - 7} =12−7= \frac{1}{2 - 7} =1−5= \frac{1}{-5}

    The limit is −15\boxed{-\frac{1}{5}}.

Part h(i)· 4 steps

  1. Substitute x=−2x=-2 into the function.

    lim⁡x→−2x3+8x2+8x+12\lim_{x \to -2} \frac{x^3 + 8}{x^2 + 8x + 12}

    Substituting x=−2x=-2 gives (−2)3+8(−2)2+8(−2)+12=−8+84−16+12=00\frac{(-2)^3 + 8}{(-2)^2 + 8(-2) + 12} = \frac{-8+8}{4-16+12} = \frac{0}{0}, which is an indeterminate form.

  2. Factorize the numerator and denominator.

    The numerator is a sum of cubes: x3+8=(x+2)(x2−2x+4)x^3 + 8 = (x+2)(x^2 - 2x + 4).
    The denominator is a quadratic: x2+8x+12=(x+2)(x+6)x^2 + 8x + 12 = (x+2)(x+6).

  3. Rewrite the limit and cancel the common factor (x+2)(x+2).

    lim⁡x→−2(x+2)(x2−2x+4)(x+2)(x+6)=lim⁡x→−2x2−2x+4x+6\lim_{x \to -2} \frac{(x+2)(x^2 - 2x + 4)}{(x+2)(x+6)} = \lim_{x \to -2} \frac{x^2 - 2x + 4}{x+6}

  4. Substitute x=−2x=-2 into the simplified expression.

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