/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Determine the following limits. ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 17

Determine the following limits. $$\lim _{x \rightarrow \infty} x^{-6}$$

Short Answer

Expert verified
Question: Calculate the limit of the function \(f(x) = x^{-6}\) as \(x\) approaches infinity. Answer: The limit of the function \(f(x) = x^{-6}\) as \(x\) approaches infinity is 0.

Step by step solution

01

Identify the function

We are given the function \(f(x) = x^{-6}\), and we need to find the limit as \(x \rightarrow \infty\).
02

Analyze the function

Now, we can analyze the given function \(f(x) = x^{-6}\) to see its behavior when \(x\) becomes very large. Since the power of \(x\) is negative, then the function gets smaller and smaller as x gets larger and larger, which indicates that the behavior of the function approaches 0 at infinity.
03

Calculate the limit

Now that we know the behavior of the function, we can directly evaluate the limit as follows: $$\lim_{x\rightarrow\infty} x^{-6} = \lim_{x\rightarrow\infty} \frac{1}{x^6}$$ As \(x\) approaches infinity, the value of the fraction \(\frac{1}{x^6}\) becomes very small since the denominator, \(x^6\), is increasing rapidly. Hence, the function approaches 0.
04

Present the result

Based on our analysis, we can conclude that the given limit is: $$\lim_{x\rightarrow\infty} x^{-6} = 0$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit at Infinity
Understanding the concept of a limit at infinity is crucial for interpreting how functions behave as their inputs grow without bound. Simply put, this limit tells us what value a function approaches as its variable, often denoted as 'x', heads towards positive or negative infinity. When we are given the expression \( \lim_{x\rightarrow\infty} x^{-6} \), our task is to determine what value \( f(x) = x^{-6} \) approaches as \( x \) becomes very large.

As \( x \) grows, the magnitude of \( x^{-6} \) becomes smaller since it's in a negative exponent form鈥攎ore on this later. A key point here is to recognize the behavior: as \( x \) increases uncontrollably, \( f(x) \) gets closer and closer to zero. This is why we conclude that \( \lim_{x\rightarrow\infty} x^{-6} = 0 \).
Negative Exponent
A negative exponent indicates division by the number raised to the corresponding positive exponent. For example, \( x^{-n} \) is equivalent to \( \frac{1}{x^n} \) for a positive integer 'n'. This inversely proportional relationship implies that as the value of \( x \) increases, the value of \( x^{-n} \) decreases.

In the context of the given exercise, \( x^{-6} \) is the same as \( \frac{1}{x^6} \). As \( x \) approaches infinity, the denominator \( x^6 \) becomes very large, making the entire fraction infinitesimally small, heading towards zero. This is a critical point of understanding for evaluating the limits involving negative exponents.
Asymptotic Behavior
The term 'asymptotic behavior' refers to the way a function behaves as it approaches a certain line or value, but never quite reaches it. It describes the end behavior of a function and is essential in studying limits. For \( x^{-6} \), as \( x \) increases, the function value gets closer to the x-axis but does not actually touch it, illustrating that the x-axis is an asymptote.

When looking at \( \lim_{x\rightarrow\infty} x^{-6} \), we evoke this concept to support the understanding that as \( x \) becomes infinitely large, the function asymptotically approaches zero, the horizontal asymptote. This approach forms a key part of the graphical interpretation of limits and is mirrored across various functions, enabling a visual and conceptual grasp of their indefinite growth or decay.

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Most popular questions from this chapter

Calculator limits Estimate the value of the following limits by creating a table of function values for \(h=0.01,0.001,\) and 0.0001 and \(h=-0.01,-0.001,\) and -0.0001. $$\lim _{h \rightarrow 0} \frac{\ln (1+h)}{h}$$

It can be shown that \(1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1,\) for \(x\) near 0 a. Illustrate these inequalities with a graph. b. Use these inequalities to evaluate \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\).

Evaluate the following limits or state that they do not exist. $$\lim _{x \rightarrow 0^{+}} \frac{x}{\ln x}$$

The limit at infinity \(\lim _{x \rightarrow \infty} f(x)=L\) means that for any \(\varepsilon>0\) there exists \(N>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad x>N.$$ Use this definition to prove the following statements. $$\lim _{x \rightarrow \infty} \frac{2 x+1}{x}=2$$

Sketching graphs of functions Sketch the graph of a function with the given properties. You do not need to find a formula for the function. $$\begin{aligned} &h(-1)=2, \lim _{x \rightarrow-1^{-}} h(x)=0, \lim _{x \rightarrow-1^{+}} h(x)=3\\\ &h(1)=\lim _{x \rightarrow 1^{-}} h(x)=1, \lim _{x \rightarrow 1^{+}} h(x)=4 \end{aligned}$$
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