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Chapter 5: Problem 38

Show that one of the limits \(\lim _{x \rightarrow 0+} f(x)\) and \(\lim _{x \rightarrow 0-} f(x)\) of the function $$ f(x)=e^{1 / x} $$ at 0 is infinite and one is finite. What can you say about the limits $$ \lim _{x \rightarrow \infty} f(x) \text { and } \lim _{x \rightarrow-\infty} f(x) ? $$

Short Answer

Expert verified
\(\lim_{x \to 0^+} f(x) = \infty\) and \(\lim_{x \to 0^-} f(x) = 0\); both \(\lim_{x \to \infty} f(x)\) and \(\lim_{x \to -\infty} f(x) = 1\).

Step by step solution

01

Analyze the behavior as x approaches 0 from the right

To find \(\lim_{x \to 0^+} f(x)\), consider \(f(x) = e^{1/x}\) as \(x\) approaches 0 from the right. When \(x\) is positive and very small, \(1/x\) is a large positive number because \(x\) is a very small positive denominator. Thus, \(e^{1/x}\) becomes extremely large.
02

Conclude the limit from the right

Based on Step 1, we conclude that as \(x\) approaches 0 from the right, \(f(x)\) approaches infinity. Therefore, \(\lim_{x \to 0^+} f(x) = \infty\).
03

Analyze the behavior as x approaches 0 from the left

To determine \(\lim_{x \to 0^-} f(x)\), analyze \(f(x) = e^{1/x}\) for values just left of 0. As \(x\) becomes a small negative number, \(1/x\) becomes a large negative number, thus \(e^{1/x}\) approaches 0 because \(e\) raised to a very large negative power approaches zero.
04

Conclude the limit from the left

Therefore, for values of \(x\) approaching 0 from the left, \(f(x)\) approaches 0. This gives us \(\lim_{x \to 0^-} f(x) = 0\).
05

Consider the limit as x approaches infinity

For \(\lim_{x \to \infty} f(x)\), as \(x\) grows larger, \(1/x\) approaches zero and \(f(x) = e^{1/x}\) approaches \(e^0 = 1\).
06

Determine the limit at infinity

Thus, \(\lim_{x \to \infty} f(x) = 1\).
07

Consider the limit as x approaches negative infinity

For \(\lim_{x \to -\infty} f(x)\), as \(x\) becomes very large negatively, \(1/x\) is a small negative number, making \(f(x) = e^{1/x}\) approach \(e^0 = 1\).
08

Determine the limit at negative infinity

This gives us the limit \(\lim_{x \to -\infty} f(x) = 1\).

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Key Concepts

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

One-Sided Limits
In calculus, one-sided limits help us understand how a function behaves as it approaches a certain point from one side only. We look at either the left side or the right side of that point. For the function \(f(x) = e^{1/x}\), we examine its behavior as \(x\) moves towards 0 from both directions:
  • From the right (\(x \to 0^+\)): As \(x\) is a small positive number, \(1/x\) becomes very large. This makes \(e^{1/x}\) increase rapidly. Therefore, the limit \(\lim_{x \to 0^+} f(x) = \infty\), indicating an infinite limit on this side.
  • From the left (\(x \to 0^-\)): As \(x\) is a small negative number, \(1/x\) becomes a large negative value. In this case, \(e^{1/x}\) diminishes to zero. Therefore, \(\lim_{x \to 0^-} f(x) = 0\), reflecting a finite limit on this side.
By understanding one-sided limits, we get a better sense of the function's behavior at critical points, even when the limits from different sides differ.
Infinite Limits
Infinite limits in calculus occur when the values of a function grow without bound as \(x\) approaches a particular point. These limits help capture scenarios where the function's value skyrockets to infinity.
With \(f(x) = e^{1/x}\), the limit \(\lim_{x \to 0^+} f(x) = \infty\) is a typical case of an infinite limit as \(x\) approaches 0 from the right. This infinite behavior emerges from the exponential function, \(e^{1/x}\), where \(1/x\) turns into an extensive positive value.
Such behavior is crucial in understanding asymmetrical properties in functions and how they diverge towards infinity on approaching certain values.
Behavior of Exponential Functions
Exponential functions like \(e^{x}\) are fundamental in calculus due to their unique properties, including their growth patterns. They exhibit rapid growth or decay, based on the exponent.
  • If the exponent is positive, \(e^x\) grows quickly.
  • If the exponent is negative, \(e^x\) rapidly approaches zero.

For \(f(x) = e^{1/x}\),
  • When \(x\) is positive, \(1/x\) is positive, leading \(e^{1/x}\) to grow.
  • When \(x\) is negative, \(1/x\) is negative, causing \(e^{1/x}\) to shrink towards zero.
Understanding these exponential behaviors is essential for determining how functions like \(f(x)\) react as their inputs change, highlighting why some limits are infinite while others remain finite.
Limits at Infinity
Limits at infinity describe the behavior of functions as their input grows very large, either positively or negatively. They help determine how functions settle down or stabilize. Consider \(f(x) = e^{1/x}\):
  • As \(x \rightarrow \infty\), \(1/x\) approaches zero from the positive side. Hence, \(f(x)\) simplifies to \(e^0 = 1\). This provides \(\lim_{x \to \infty} f(x) = 1\).
  • As \(x\rightarrow -\infty\), \(1/x\) also trends towards zero, though from the negative side. Still, \(f(x)\) evaluates to \(e^0 = 1\). This means \(\lim_{x \to -\infty} f(x) = 1\).
At both ends of the spectrum, the function \(f(x)\) approaches 1, indicating it stabilizes to this value regardless of the direction of infinity. Knowing limits at infinity is invaluable in assessing long-term behavior of functions and their eventual convergence.

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Most popular questions from this chapter

Let \(f\) be a uniformly continuous function on a set \(E\). Show that if \(\left\\{x_{n}\right\\}\) is a Cauchy sequence in \(E\) then \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence in \(f(E)\). Show that this need not be true if \(f\) is continuous but not uniformly continuous.

Show that $$ \lim _{x \rightarrow x_{0}} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow x_{0}} f(x)} $$ could be false if \(\lim _{x \rightarrow x_{0}} f(x)=0\) and definitely is false if \(\lim _{x \rightarrow x_{0}} f(x)<0\)

Can the range of an increasing function on the interval \([0,1]\) consist only of rational numbers? Can it consist only of irrational numbers?

Let \(f\) be continuous in a neighborhood \(U\) of the point \(x_{0}\). If \(f(x)<\beta\) for all \(x \in U \backslash\left\\{x_{0}\right\\}\), prove that \(f\left(x_{0}\right) \leq \beta .\) Show by example that we cannot conclude \(f\left(x_{0}\right)<\beta\).

Give an example of a function defined only on the rationals and continuous at each point in its domain and yet does not have an absolute maximum.
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