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Evaluate the limit of each of the following: (i) limx->2 (3-x)/(x+2) (ii) limx->1 (5-x)/(x+3) (iii) limx->inf (cbrt(x)-3x+2)/(3x+x^(1/3)+2) (iv) limx->1 (x2-2)/(x-3)

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Answers

4 parts
a

14\frac{1}{4}

b
1
c

13\frac{1}{3}

d

13\frac{1}{3}

Step-by-step solution

9 steps

Part a· 2 steps

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

    lim⁡x→23−xx+2=3−22+2\lim_{x \to 2} \frac{3-x}{x+2} = \frac{3-2}{2+2}

  2. Simplify the expression.

    14\frac{1}{4}

    The limit is 14\boxed{\frac{1}{4}}.

Part b1

  1. Substitute x=1x=1 into the expression.

    lim⁡x→15−xx+3=5−11+3\lim_{x \to 1} \frac{5-x}{x+3} = \frac{5-1}{1+3}

  2. Simplify the expression.

    44=1\frac{4}{4} = 1

    The limit is 1\boxed{1}.

Part c

13\frac{1}{3}

  1. Identify the highest power of xx in the denominator, which is xx.

    Divide both the numerator and the denominator by xx.

    lim⁡x→∞x3−3x+233x+x1/3+2=lim⁡x→∞x3−3x+23x3x+x1/3+2x\lim_{x \to \infty} \frac{\sqrt[3]{x^3 - 3x + 2}}{3x+x^{1/3}+2} = \lim_{x \to \infty} \frac{\frac{\sqrt[3]{x^3 - 3x + 2}}{x}}{\frac{3x+x^{1/3}+2}{x}}

  2. Rewrite the terms by moving xx inside the cube root as x3x^3 and simplifying.

    lim⁡x→∞x3−3x+2x333xx+x1/3x+2x=lim⁡x→∞1−3x2+2x333+1x2/3+2x\lim_{x \to \infty} \frac{\sqrt[3]{\frac{x^3 - 3x + 2}{x^3}}}{\frac{3x}{x}+\frac{x^{1/3}}{x}+\frac{2}{x}} = \lim_{x \to \infty} \frac{\sqrt[3]{1 - \frac{3}{x^2} + \frac{2}{x^3}}}{3+\frac{1}{x^{2/3}}+\frac{2}{x}}

  3. Apply the limit.

    As x→∞x \to \infty, terms like cxn\frac{c}{x^n} approach 0.

    1−0+033+0+0=133=13\frac{\sqrt[3]{1 - 0 + 0}}{3+0+0} = \frac{\sqrt[3]{1}}{3} = \frac{1}{3}

    The limit is 13\boxed{\frac{1}{3}}.

Part d

13\frac{1}{3}

  1. Substitute x=1x=1 into the expression.

    lim⁡x→1x2−2x2−x−3=12−212−1−3\lim_{x \to 1} \frac{x^2-2}{x^2-x-3} = \frac{1^2-2}{1^2-1-3}

  2. Simplify the expression.

    1−21−1−3=−1−3=13\frac{1-2}{1-1-3} = \frac{-1}{-3} = \frac{1}{3}

    The limit is 13\boxed{\frac{1}{3}}.

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