Question

multiple-choice problems

1. evaluate \\(\lim\limits_{x\to -\infty} \frac{4x^3 + 5x^2 - 1}{2x^3 - 7}\\).

a. -2
b. 2
c. -\infty
d. \infty
e. answer not given

1. consider \\(f(x) = \frac{1}{x^3 - 2}\\). evaluate \\(\lim\limits_{x\to \sqrt3{2}} f(x)\\).

a. 0
b. 1/2
c. \infty
d. -\infty
e. answer not given

Explanation:

Step1: Divide by highest power $x^3$

$$\lim_{x \to -\infty} \frac{4x^3 + 5x^2 - 1}{2x^3 - 7} = \lim_{x \to -\infty} \frac{4 + \frac{5}{x} - \frac{1}{x^3}}{2 - \frac{7}{x^3}}$$

Step2: Evaluate limits of fractions

As $x \to -\infty$, $\frac{5}{x} \to 0$, $\frac{1}{x^3} \to 0$, $\frac{7}{x^3} \to 0$.
$$\lim_{x \to -\infty} \frac{4 + 0 - 0}{2 - 0} = \frac{4}{2}$$

Step3: Simplify the result

$$\frac{4}{2} = 2$$

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Step1: Substitute $x=(\sqrt{2})^+$ into denominator

First, calculate $x^2 - 1$ when $x \to (\sqrt{2})^+$: $(\sqrt{2})^2 - 1 = 2 - 1 = 1$, and as $x$ approaches $\sqrt{2}$ from the right, $x^2 - 1$ approaches 0 from the positive side.

Step2: Evaluate the limit of $f(x)$

$$\lim_{x \to (\sqrt{2})^+} \frac{1}{x^2 - 1} = \lim_{x \to (\sqrt{2})^+} \frac{1}{\text{small positive number}} = \infty$$

Answer:

1. B. 2
2. C. $\infty$