Chapter 1: Problem 38
Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} x^{1 / x} $$
Short Answer
Expert verified
The limit is 0.
Step by step solution
01
Recognize the Limit Form
The expression \( x^{1/x} \) as \( x \to 0^+ \) is an indeterminate form of the type \( 0^0 \). This indicates that we may need to use logarithms or another method to evaluate the limit.
02
Apply Logarithms
To simplify \( x^{1/x} \), take the natural logarithm: \( \ln(y) = \ln(x^{1/x}) = \frac{1}{x} \ln(x) \). Here, we have \( y = x^{1/x} \), and \( \ln(y) \) now becomes our focus.
03
Evaluate the New Limit
Find the limit \( \lim_{x \to 0^+} \frac{\ln(x)}{x} \). As \( x \to 0^+ \), \( \ln(x) \to -\infty \) and \( x \to 0^+ \), indicating that we have an indeterminate form \( \frac{-\infty}{0^+} \).
04
Use L'Hôpital's Rule
Since \( \frac{\ln(x)}{x} \) is an indeterminate form of type \( \frac{-\infty}{0^+} \), apply L'Hôpital's Rule: differentiate the numerator and the denominator. The derivative of \( \ln(x) \) is \( \frac{1}{x} \), and the derivative of \( x \) is \( 1 \).
05
Evaluate the New Limit after Differentiation
Now find \( \lim_{x \to 0^+} \frac{1/x}{1} = \lim_{x \to 0^+} \frac{1}{x} = -\infty \). Thus, \( \ln(y) \to -\infty \) as \( x \to 0^+ \).
06
Interpret Result
Since \( \ln(y) \to -\infty \), the expression \( y = e^{-\infty} = 0 \). Therefore, the original limit is \( \lim_{x \to 0^+} x^{1/x} = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
Indeterminate forms are expressions involving limits that do not initially provide clear information about the resulting value. These forms can include expressions like \( 0/0, \ \infty / \infty, \ 0^0,\ 1^{\infty}, \ 0 \times \infty,\) etc. These do not immediately resolve to a single value and usually require additional manipulation to evaluate.
In the original exercise, the expression \( x^{1/x} \) as \( x \to 0^+ \) becomes an indeterminate form \( 0^0 \). This specific form signals a need for us to rewrite the expression in a way that can offer clarity. Utilizing techniques such as logarithms or changing how we approach the limit can allow these forms to be evaluated correctly.
Recognizing and appropriately handling indeterminate forms is crucial since directly substituting the approaching value won't work as expected. Therefore, handling them with care and employing the right mathematical tools is a fundamental skill when evaluating limits.
In the original exercise, the expression \( x^{1/x} \) as \( x \to 0^+ \) becomes an indeterminate form \( 0^0 \). This specific form signals a need for us to rewrite the expression in a way that can offer clarity. Utilizing techniques such as logarithms or changing how we approach the limit can allow these forms to be evaluated correctly.
Recognizing and appropriately handling indeterminate forms is crucial since directly substituting the approaching value won't work as expected. Therefore, handling them with care and employing the right mathematical tools is a fundamental skill when evaluating limits.
Natural Logarithms
Natural logarithms are a core tool for simplifying expressions, especially when evaluating limits involves exponential or power functions. The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \).
In the problem of dealing with \( x^{1/x} \), we took the natural logarithm to transform the expression into a linear form. This step is pivotal because the natural logarithm possesses properties that simplify complex equations. For instance, taking logarithms allows power expressions to be dealt with multiplication instead, since \( \ln(x^a) = a \cdot \ln(x) \).
Through this method, \( \ln(x^{1/x}) \) becomes \( \frac{1}{x} \ln(x) \), which can further undergo techniques like L'Hôpital's rule for limit evaluation. Natural logarithms, thus, pave the way for transforming and simplifying the original function, making an indeterminate form approachable.
In the problem of dealing with \( x^{1/x} \), we took the natural logarithm to transform the expression into a linear form. This step is pivotal because the natural logarithm possesses properties that simplify complex equations. For instance, taking logarithms allows power expressions to be dealt with multiplication instead, since \( \ln(x^a) = a \cdot \ln(x) \).
Through this method, \( \ln(x^{1/x}) \) becomes \( \frac{1}{x} \ln(x) \), which can further undergo techniques like L'Hôpital's rule for limit evaluation. Natural logarithms, thus, pave the way for transforming and simplifying the original function, making an indeterminate form approachable.
Limit Evaluation
Limit evaluation is the process of finding the limit of a function as it approaches a particular value. When evaluating limits, it's not always straightforward, especially when dealing with indeterminate forms.
In our task, once we acknowledge that the expression forms an indeterminate type, we transform it using logarithms and compute the new limit of \( \lim_{x \to 0^+} \frac{\ln(x)}{x} \). This involves considering the behavior of \( \ln(x) \to -\infty \) while \( x \to 0^+ \).
To resolve this, applying L'Hôpital's rule, which is used for simplifying limits of indeterminate forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), becomes appropriate. Differentiating the numerator and the denominator separately gives new insights. Finally, after differentiating, the limit \( \lim_{x \to 0^+} \frac{1}{x} \) turns out to be \(-\infty\), thereby guiding the interpretation that \( y = e^{-\infty} = 0 \).
Evaluating limits demands a combination of algebraic manipulation and calculus techniques to elucidate the behavior of functions around points of interest, providing a comprehensive understanding of how to approach complex expressions.
In our task, once we acknowledge that the expression forms an indeterminate type, we transform it using logarithms and compute the new limit of \( \lim_{x \to 0^+} \frac{\ln(x)}{x} \). This involves considering the behavior of \( \ln(x) \to -\infty \) while \( x \to 0^+ \).
To resolve this, applying L'Hôpital's rule, which is used for simplifying limits of indeterminate forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), becomes appropriate. Differentiating the numerator and the denominator separately gives new insights. Finally, after differentiating, the limit \( \lim_{x \to 0^+} \frac{1}{x} \) turns out to be \(-\infty\), thereby guiding the interpretation that \( y = e^{-\infty} = 0 \).
Evaluating limits demands a combination of algebraic manipulation and calculus techniques to elucidate the behavior of functions around points of interest, providing a comprehensive understanding of how to approach complex expressions.
