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Chapter 2: Problem 22

Determine the following limits. $$\lim _{x \rightarrow-\infty} 2 x^{-8}$$

Short Answer

Expert verified
Answer: The limit of the function is 0.

Step by step solution

01

Identify the form of the function

The given function is of the form $$f(x) = 2x^{-8}.$$ As x approaches negative infinity, the exponentiation part of the function is \(x^{-8}\), which gets very small because the exponent is negative.
02

Find the limit

To find the limit, we need to consider the behavior of the function as x approaches negative infinity. For very large negative x values, the exponentiation part of the function \(x^{-8}\) will get very close to 0, as \(x^{-8} = \frac{1}{x^8}\), and, when x is negative infinity, the denominator becomes infinitely large making the term very small. Now, consider the given function: $$\lim_{x \to -\infty} 2x^{-8}.$$ Since the exponentiation part will approach 0, we can rewrite the limit as: $$\lim_{x \to -\infty} \frac{2}{x^8}.$$ As x approaches negative infinity, the fraction becomes \(\frac{2}{(\infty)^8}\). Therefore, the limit is: $$\lim_{x \to -\infty} \frac{2}{x^8} = 0.$$
03

State the conclusion

The limit of the function $$2x^{-8}$$ as x approaches negative infinity is 0. That is, $$\lim _{x \rightarrow-\infty} 2 x^{-8} = 0.$$

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Most popular questions from this chapter

A sequence is an infinite, ordered list of numbers that is often defined by a function. For example, the sequence \(\\{2,4,6,8, \ldots\\}\) is specified by the function \(f(n)=2 n\), where \(n=1,2,3, \ldots .\) The limit of such a sequence is \(\lim _{n \rightarrow \infty} f(n)\), provided the limit exists. All the limit laws for limits at infinity may be applied to limits of sequences. Find the limit of the following sequences, or state that the limit does not exist. \(\left\\{0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots .\right\\},\) which is defined by \(f(n)=\frac{n-1}{n},\) for \(n=1,2,3, \ldots\)

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