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Prove that $$\lim _{x \rightarrow \infty} f(x)=\lim _{t \rightarrow 0^{+}} f(1 / t)$$ and $$\lim _{x \rightarrow-\infty} f(x)=\lim _{t \rightarrow 0^{-}} f(1 / t)$$ if these limits exist.

Short Answer

Expert verified
Substitute \( x = \frac{1}{t} \) and relate the expressions through the behavior at zero.

Step by step solution

01

Understand the Limit Definition

The definition of a limit for a function \( f(x) \) as \( x \) approaches a value \( a \) is that the values of \( f(x) \) get arbitrarily close to \( L \) as \( x \) approaches \( a \). In this problem, we are dealing with infinity, so the function behavior is analyzed as \( x \rightarrow \infty \) or \( x \rightarrow -\infty \).
02

Rewriting Limits with Substitution

To prove the given limits, consider a substitution \( x = \frac{1}{t} \). As \( x \rightarrow \infty \), \( \frac{1}{x} \rightarrow 0^{+} \), and as \( x \rightarrow -\infty \), \( \frac{1}{x} \rightarrow 0^{-} \). Thus, the task is to show the relation between the limits in terms of \( t \).
03

Analyze the Limit \( \lim_{x \rightarrow \infty} f(x) \)

Given \( x = \frac{1}{t} \), when \( x \rightarrow \infty \), it implies \( t \rightarrow 0^{+} \). Hence, it follows that \( \lim_{x \rightarrow \infty} f(x) = \lim_{t \rightarrow 0^{+}} f(\frac{1}{t}) \) when the limits exist, because both expressions represent the same approach path for \( f(x) \).
04

Analyze the Limit \( \lim_{x \rightarrow -\infty} f(x) \)

Similarly, for \( x = \frac{1}{t} \), \( x \rightarrow -\infty \) implies \( t \rightarrow 0^{-} \). Thus, \( \lim_{x \rightarrow -\infty} f(x) = \lim_{t \rightarrow 0^{-}} f(\frac{1}{t}) \). Here again, both expressions represent the same approach for \( f(x) \), confirming they are equivalent when the limits exist.

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Key Concepts

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

Existence of Limits
The concept of limits plays a crucial role in calculus, as it provides a way to understand the behavior of functions as they approach specific points or grow indefinitely. To say that a limit exists means that as the variable approaches a certain point, the function approaches a particular value. In mathematical terms, for the limit of a function \( f(x) \) as \( x \rightarrow a \) to exist, the function values must get arbitrarily close to a number \( L \).

In the context of the provided exercise, we are dealing with limits at infinity. This implies examining the behavior of the function as \( x \rightarrow \infty \) or \( x \rightarrow -\infty \). If such limits exist, it indicates that despite \( x \) becoming indefinitely large or small, \( f(x) \) approaches a consistent value. This property is essential to ensure that transitions between representations like \( f(x) \) and \( f(1/t) \) hold true as explained in the problem statement.
  • Existence of limits assures consistency in function behavior.
  • It allows us to represent functions differently while remaining equivalent.
  • Understanding this ensures solving problems with infinite behavior correctly.
Infinity in Calculus
Infinity is a unique and sometimes tricky concept in calculus. It represents values larger than any finite number but not a number itself. As \( x \rightarrow \infty \), we analyze how a function behaves as \( x \) continues to grow without bounds. Similarly, \( x \rightarrow -\infty \) examines behavior as \( x \) decreases indefinitely into the negative.

In the exercise provided, expressing limits as \( x \rightarrow \infty \) and \( x \rightarrow -\infty \) helps to understand the asymptotic behavior of the function. This means we try to establish how the function behaves as \( x \) moves towards these infinite points. If \( f(x) \) approaches a specific value, we can say the limit exists in infinite terms.
  • Infinity is not a number, but a concept indicating unbounded growth.
  • In calculus, infinite limits help us understand long-term behavior.
  • By considering infinity, we deduce the asymptotic trend of functions.
Substitution in Limits
Substitution is a powerful technique to evaluate limits, especially those involving complex transformations or infinite behavior. The exercise demonstrates the substitution \( x = \frac{1}{t} \), which helps transition the problem from an infinite limit of \( x \) to a finite limit as \( t \rightarrow 0^+ \) or \( t \rightarrow 0^- \). This equivalence utilizes the idea that if \( x \rightarrow \infty \), then \( \frac{1}{x} \rightarrow 0^+ \) and vice-versa.

This method allows us to harness more familiar finite limit techniques to solve seemingly complex problems. Recognizing that \( f(x) \) and \( f(1/t) \) describe the same function behavior means you can seamlessly shift perspectives to simplify evaluation.
  • Substitution ties infinite limits to finite, approachable forms.
  • It helps bridge different representations of the same problem.
  • By simplifying limits via substitution, complex problems become manageable.

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