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Chapter 2: Problem 43

Prove that $$\lim _{x \rightarrow 0^{+}} \ln x=-\infty$$

Short Answer

Expert verified
As \( x \rightarrow 0^{+} \), \( \ln x \) decreases indefinitely, thus \( \lim _{x \rightarrow 0^{+}} \ln x = -\infty \).

Step by step solution

01

Understand the Limit Expression

The expression \( \lim _{x \rightarrow 0^{+}} \ln x = -\infty \) means we need to show that as \( x \) approaches 0 from the positive side, the value of \( \ln x \) decreases without bound (becomes infinitely negative). This involves evaluating how \( \ln x \) behaves as \( x \) gets closer to 0.
02

Analyze the Behavior of the Natural Logarithm Function

The natural logarithm function \( \ln x \) is only defined for positive values of \( x \). As \( x \) approaches positive values closer to 0, \( \ln x \) drops sharply. Observationally, we can note that \( \ln x \) becomes more and more negative as \( x \) decreases towards 0.
03

Argue using the Properties of Natural Logarithm

For \( 0 < x < 1 \), the value of \( \ln x \) is negative because the natural log of numbers between 0 and 1 is negative. More formally, notice that \( \ln(x) \) is the inverse of the exponential function; therefore as \( x \rightarrow 0^{+} \), \( e^{\ln x} \rightarrow 0 \). Both \( \ln(1) = 0 \) and as \( x \) moves closer to 0, the slope of \( \ln x \) causes it to drop infinitely negatively.
04

Apply a Formal Limit Approach

Consider any large negative number \( N \). We need to show that there exists a positive \( \delta \) such that for all \( 0 < x < \delta \), \( \ln x < N \). Take \( \delta = e^N \). For any \( 0 < x < \delta \), we have \( 0 < x < e^N \) implying \( \ln x < N \), since \( \ln(e^N) = N \). This confirms that \( \ln x \rightarrow -\infty \) as \( x \rightarrow 0^{+} \).

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Key Concepts

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

Natural Logarithm
The natural logarithm, represented as \( \ln x \), is a fundamental mathematical function widely used in calculus and many fields of science and engineering. Let's break this down further to understand its special characteristics:
  • Definition and Domain: The natural logarithm is only defined for positive values of \( x \). This means \( \ln x \) can be used only when \( x > 0 \).
  • Inverse Property: The natural logarithm is the inverse of the exponential function (i.e., \( e^{y} = x \) implies \( y = \ln x \)). This relationship is key to understanding the behavior of logarithmic functions.
  • Slope and Growth: For values of \( x \) between 0 and 1, \( \ln x \) is negative, and as \( x \) decreases towards zero, \( \ln x \) decreases rapidly. This results in a steep negative slope as you move leftward towards zero.
Each of these aspects plays a critical role in exploring limits and understanding the behavior of functions as \( x \) changes.
Limits at Infinity
Calculus often deals with limits, which describe the behavior of a function as \( x \) approaches a certain value or infinity. Let's dive into what limits at infinity entail:
  • Concept: The notation \( \lim_{x \to \infty} \) or \( \lim_{x \to 0^{+}} \) indicates we are interested in the behavior of a function as \( x \) becomes very large, or very close to zero, respectively.
  • Natural Logarithm and Infinity: For the natural logarithm \( \ln x \), as \( x \to \infty \), \( \ln x \to \infty \). Conversely, as \( x \to 0^{+} \), \( \ln x \) goes to \(-\infty\).
Consider the limit \( \lim_{x \to 0^{+}} \ln x = -\infty \). This implies that as \( x \) gets tinier and tinier, closer to zero from the positive side, the values of \( \ln x \) become hugely negative, without bound.
Behavior as x Approaches Zero
Understanding how functions behave as \( x \) approaches zero is essential in calculus. It gives insight into the limiting behavior and helps solve many mathematical problems.
  • Approaching Zero from Positive Side: When we examine the limit \( \lim_{x \to 0^{+}} \ln x \), we focus on \( x \) reducing towards 0 from the positive side.
  • Function Behavior: As \( x \) approaches zero, \( \ln x \) drops sharply towards negative infinity since the natural log of numbers less than one is negative.
  • Mathematical Proof: By picking any large negative number \( N \) and setting \( \delta = e^N \), we prove that for any \( x < \delta \), \( \ln x \) becomes smaller than \( N \). Hence, it drops beyond any bound as \( x \to 0^{+} \).
These steps illustrate how \( \ln x \) behaves dramatically as it approaches zero, providing a clear mathematical foundation for its limit behavior.

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Most popular questions from this chapter

Determine the infinite limit. $$\lim _{x \rightarrow \pi^{-}} \cot x$$

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion \(s=2 \sin \pi t+3 \cos \pi t,\) where \(t\) is measured in seconds. (a) Find the average velocity during each time period: $$\begin{array}{ll}{\text { (i) }[1,2]} & {\text { (ii) }[1,1.1]} \\ {\text { (iii) }} & {[1,1.01]} & {[\text { iv) }[1,1.001]}\end{array}$$ (b) Estimate the instantaneous velocity of the particle when t= 1.

J Use the graph of the function \(f(x)=1 /\left(1+e^{1 / x}\right)\) to state the value of each limit, if it exists. If it does not exist, explain why. $${ (a) }\lim _{x \rightarrow 0^{-}} f(x) \quad \text { (b) } \lim _{x \rightarrow 0^{+}} f(x) \quad \text { (c) } \lim _{x \rightarrow 0} f(x)$$

$$ \begin{array}{l}{\text { Make a rough sketch of the curve } y=x^{n}(n \text { an integer })} \\ {\text { for the following five cases: }}\end{array} $$ $$ \begin{array}{ll}{\text { (i) } \mathrm{n}=0} & {\text { (ii) } \mathrm{n}>0, \mathrm{n} \text { odd }} \\ {\text { (iii) } \mathrm{n}>0, \mathrm{n} \text { even }} & {\text { (iv) } \mathrm{n}<0, \mathrm{n} \text { odd }}\end{array} $$ $$ \begin{array}{l}{\text { (v) } \mathrm{n}<0, \text { n even }} \\ {\text { Then use these sketches to find the following limits. }}\end{array} $$ (a) $$ \lim _{x \rightarrow \mathbb{u}^{+}} x^{n} \quad \text { (b) } \lim _{x \rightarrow 0^{-}} x^{n} $$ (c) $$ \lim _{x \rightarrow \infty} x^{n} \quad \text { (d) } \lim _{x \rightarrow-\infty} x^{n} $$

$$ \begin{array}{c}{\text { Use a graph to find a number } N \text { such that }} \\ {\text { If } \quad x>N \quad \text { then } \quad\left|\frac{3 x^{2}+1}{2 x^{2}+x+1}-1.5\right|<0.05}\end{array} $$
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