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Chapter 1: Q. 54 (page 149)

Calculate each limit in Exercises 35鈥�80.
$\underset{x 鈫� 0^{+}}{l i m} \frac{x^{- 3}}{x^{2} - x^{- 1}}$

Short Answer

Expert verified

The limit is infinity.

Step by step solution

01

Step 1. Given information

The given expression is $\underset{x 鈫� 0^{+}}{l i m} \frac{x^{- 3}}{x^{2} - x^{- 1}}$

02

Step 2. Calculation

The limit is calculated as below,

\underset{x 鈫� 0^{+}}{l i m} \frac{x^{- 3}}{x^{2} - x^{- 1}} = \underset{x 鈫� 0^{+}}{l i m} \frac{\frac{1}{x^{3}}}{x^{2} - \frac{1}{x}} = \underset{x 鈫� 0^{+}}{l i m} \frac{\frac{1}{x^{3}}}{\frac{x^{3} - 1}{x}} = \underset{x 鈫� 0^{+}}{l i m} \frac{1}{x^{2} (x^{3} - 1)} = \frac{1}{0^{2} (0^{3} - 1)} = 鈭�

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Most popular questions from this chapter

$\lim_{x 鈫� 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$

In Exercises 39鈥�44, use Theorem 1.16 and left and right limits to determine whether each function f is continuous at its break point(s). For each discontinuity of f, describe the type of discontinuity and any one-sided discontinuity.

$f (x) = \{\begin{matrix} x^{2}, i f x < 2 \\ 4, i f x = 2 \\ 2 x + 1, i f x > 2 \end{matrix} .$

Use algebra to find the largest possible value of 未 or smallest possible value of N that makes each implication true. Then verify and support your answers with labeled graphs.

$If x > N, then |\frac{1}{x^{2}} 鈭� 0| < 0.001$

State what it means for a functionf to be continuous at a point x = c, in terms of the delta鈥揺psilon definition of limit.

$\lim_{x 鈫� 0} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$

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