Chapter 7: Problem 52
Find the limits in Exercises \(51-66\) $$\lim _{x \rightarrow 1^{+}} x^{1 /(x-1)}$$
Short Answer
Expert verified
The limit is \( e \).
Step by step solution
01
Rewriting the Function
Let's start by rewriting the function to make it easier to work with. We have \( x^{1/(x-1)} \), and to simplify it, we'll take the natural logarithm of the function. This gives us \( \ln(y) = \frac{\ln(x)}{x-1} \), where \( y = x^{1/(x-1)} \).
02
Applying L'Hôpital's Rule
As \( x \to 1^+ \), both the numerator and denominator in \( \frac{\ln(x)}{x-1} \) approach 0, creating an indeterminate form \( \frac{0}{0} \). L'Hôpital's Rule can resolve this by differentiating the numerator and denominator: \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \) and \( \frac{d}{dx}[x-1] = 1 \). Applying L'Hôpital's Rule, we get \( \lim_{x \to 1^+} \frac{1/x}{1} = \lim_{x \to 1^+} \frac{1}{x} \).
03
Evaluating the Limit
Now evaluate the limit: \( \lim_{x \to 1^+} \frac{1}{x} = 1 \). Therefore, \( \lim_{x \to 1^+} \ln(y) = 1 \).
04
Converting Back from a Logarithm
Since \( \ln(y) = 1 \), it follows that \( y = e^1 = e \). Since \( y = x^{1/(x-1)} \), the original limit \( \lim_{x \to 1^+} x^{1/(x-1)} = e \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus that helps us solve limits that initially seem indeterminate, like the ratio \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It works by allowing us to differentiate the numerator and the denominator of a fraction separately and then take the limit of the resulting expression. However, it's important to remember that both the original numerator and denominator have to approach 0 or infinity for the rule to be applicable.
Here's how it works:
Here's how it works:
- You identify the indeterminate form, primarily \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
- Differentiation of both the numerator and the denominator has to be performed individually.
- Then, you take the limit of the resulting expression to find the answer.
Natural Logarithm
The natural logarithm, often denoted as \( \ln(x) \), is a special logarithm that uses the base \( e \), which is approximately 2.71828. This logarithm often appears in calculus and natural growth processes because of its intrinsic properties with respect to exponential functions. When we say \( \ln(e) = 1 \), it simply means that the power we need to raise \( e \) to get \( e \) is 1.
The natural logarithm is useful in calculus for transforming exponentials into linear functions, making them easier to handle. For example, when simplifying the expression \( x^{1/(x-1)} \) in our exercise using natural logs, the exponent becomes a well-behaved fraction \( \frac{\ln(x)}{x-1} \).
Key characteristics of the natural logarithm include:
The natural logarithm is useful in calculus for transforming exponentials into linear functions, making them easier to handle. For example, when simplifying the expression \( x^{1/(x-1)} \) in our exercise using natural logs, the exponent becomes a well-behaved fraction \( \frac{\ln(x)}{x-1} \).
Key characteristics of the natural logarithm include:
- The domain is \( x>0 \), meaning it can only take positive inputs.
- It increases slowly and is defined for all positive \( x \).
- The derivative of \( \ln(x) \) is \( \frac{1}{x} \), a frequently used property in differentiation.
Indeterminate Forms
In calculus, an indeterminate form arises when the limit of a function cannot be directly evaluated because it initially doesn't lead to a definitive answer. Common types include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and more. These expressions are 'indeterminate' because they don’t imply a clear numeric result without further analysis.
Our exercise presented an indeterminate form when evaluating \( \lim_{x \to 1^+} \frac{\ln(x)}{x-1} \), as both \( \ln(x) \) and \( x-1 \) approach zero. Here L'Hôpital's Rule comes in handy, but the first step in addressing indeterminate forms is recognizing them.
To deal with indeterminate forms:
Our exercise presented an indeterminate form when evaluating \( \lim_{x \to 1^+} \frac{\ln(x)}{x-1} \), as both \( \ln(x) \) and \( x-1 \) approach zero. Here L'Hôpital's Rule comes in handy, but the first step in addressing indeterminate forms is recognizing them.
To deal with indeterminate forms:
- Identify the form to confirm it's truly indeterminate, not just undefined.
- Apply mathematical techniques like L'Hôpital's Rule, algebraic manipulation, or series expansion, to resolve the issue.
- Evaluate the resulting expression to find the limit.