Chapter 8: Problem 32
Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 1}\left(\frac{1}{x-1}-\frac{x}{\ln x}\right) $$
Short Answer
Expert verified
The limit is \(-\infty\).
Step by step solution
01
Plug in the Limit Value
First, substitute \( x = 1 \) into the expression \( \frac{1}{x-1}-\frac{x}{\ln x} \). This gives us \( \frac{1}{1-1} - \frac{1}{\ln 1} = \frac{1}{0} - \frac{1}{0} \), which is an indeterminate form \( \frac{\infty - \infty}{0} \).
02
Combine the Fractions
To use l'Hôpital's Rule, we need a single fraction. Start by finding a common denominator for \( \frac{1}{x-1} \) and \( \frac{x}{\ln x} \), which is \( (x-1)\ln x \). Then rewrite the expression: \( \frac{\ln x - x(x-1)}{(x-1)\ln x} \).
03
Simplify the Numerator
Simplify the numerator: \( \ln x - x(x-1) = \ln x - x^2 + x \). Thus, the expression becomes \( \frac{\ln x - x^2 + x}{(x-1)\ln x} \).
04
Apply l'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \) at \( x=1 \), apply l'Hôpital's Rule. Differentiate the numerator to get \( \frac{d}{dx}(\ln x - x^2 + x) = \frac{1}{x} - 2x + 1 \). Differentiate the denominator to get \( \frac{d}{dx}((x-1)\ln x) = \ln x + \frac{x-1}{x} \).
05
Re-Evaluate the Limit
Now evaluate the limit: \( \lim_{x \to 1} \frac{\frac{1}{x} - 2x + 1}{\ln x + \frac{x-1}{x}} \). Substituting \( x = 1 \) again gives \( \frac{1-2+1}{0+0} = \frac{0}{0} \). This is another indeterminate form.
06
Apply l'Hôpital’s Rule Again
Apply l'Hôpital's Rule a second time. Differentiate the numerator \( \frac{1}{x} - 2x + 1 \) to get \( -\frac{1}{x^2} - 2 \). Differentiate the denominator \( \ln x + \frac{x-1}{x} \) to obtain \( \frac{1}{x} - \frac{1}{x^2} \).
07
Evaluate the New Limit
Now find \( \lim_{x \to 1} \frac{-\frac{1}{x^2} - 2}{\frac{1}{x} - \frac{1}{x^2}} \). Substitute \( x = 1 \) into the expression, which results in \( \frac{-1}{0} \), indicating the limit is approaching negative infinity.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
In calculus, when evaluating limits, you may encounter expressions that don't provide a clear-cut numerical value at first glance. These are called *indeterminate forms*. Examples of indeterminate forms include:
- \( \frac{0}{0} \)
- \( \frac{\infty}{\infty} \)
- \( \infty - \infty \)
- \( 0 \times \infty \)
- \( 0^0 \)
- \( 1^\infty \)
- \( \infty^0 \)
Limits in Calculus
Limits are foundational in calculus, representing the value a function approaches as the input gets arbitrarily close to a particular point. Understanding limits is key to grasping more complicated topics in calculus, such as derivatives and integrals.
Here are a few essential concepts about limits:
Here are a few essential concepts about limits:
- **Limit Notation:** The notation \( \lim_{x \to a} f(x) = L \) means that as \( x \) approaches \( a \), the function \( f(x) \) gets arbitrarily close to \( L \).
- **Existence of Limits:** A limit \( \lim_{x \to a} f(x) \) exists if the values of \( f(x) \) get closer to a single finite number as \( x \) approaches \( a \) from both sides.
- **Infinite Limits:** When a function increases or decreases without bound near a point, the limit can be infinite, represented as \( \pm \infty \).
Differentiation
Differentiation is a fundamental operation in calculus that deals with finding the derivative of a function. A derivative represents the rate at which a function is changing at any given point and is symbolically denoted as \( f'(x) \) or \( \frac{df}{dx} \).
The basic rules of differentiation include:
The basic rules of differentiation include:
- **Power Rule:** \( \frac{d}{dx} x^n = nx^{n-1} \)
- **Product Rule:** \( \frac{d}{dx} (uv) = u'v + uv' \)
- **Quotient Rule:** \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \)
- **Chain Rule:** \( \frac{d}{dx} f(g(x)) = f'(g(x))g'(x) \)