Chapter 3: Problem 26
Rewrite the indeterminate form of type \(0 \cdot \infty\) as either type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty} .\) Use L'H么pital's Rule to evaluate the limit. $$ \lim _{x \rightarrow 0^{+}}\left(4 x^{2}\right)(\ln x) $$
Short Answer
Expert verified
The limit is 0.
Step by step solution
01
Identify the Indeterminate Form
The given limit is \( \lim_{x \rightarrow 0^{+}} (4x^2)(\ln x) \). As \(x\) approaches \(0^+\), \(4x^2\) approaches 0 and \(\ln x\) approaches \(-\infty\), thus the expression takes the form \(0 \cdot (-\infty)\), which is indeterminate.
02
Convert to \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) Form
To apply L'H么pital's Rule, we need to rewrite the expression as a fraction. Let's rewrite \((4x^2)(\ln x)\) as \(\frac{\ln x}{1/(4x^2)}\). Now as \(x\) approaches \(0^+\), \(\ln x\) approaches \(-\infty\) and \(1/(4x^2)\) approaches \(+\infty\), which creates an indeterminate form of type \(\frac{-\infty}{\infty}\).
03
Apply L'H么pital's Rule
Since the limit \(\frac{-\infty}{\infty}\) is indeterminate, we can apply L'H么pital's Rule which states the limit of a fraction is equal to the limit of its derivatives, provided they exist. Differentiating the numerator \(\ln x\) gives \(\frac{1}{x}\) and differentiating the denominator \(1/(4x^2)\) using the quotient rule gives \(-\frac{2}{x^3}\).
04
Simplify and Evaluate the Limit
Now evaluate the limit: \[ \lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{2}{x^3}} = \lim_{x \rightarrow 0^{+}} \frac{x^3}{-2x} = \lim_{x \rightarrow 0^{+}} \frac{x^2}{-2} \]. As \(x\) approaches \(0^+\), \(x^2\) approaches 0, resulting in the overall limit being 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms in Calculus
Indeterminate forms appear in calculus when the substitution of a limit value into a function does not yield a clear answer. Common forms like
0/0, 鈭/鈭, 0鈰呪垶, 鈭炩垝鈭, and 0鈦 can be problematic. When faced with such forms, direct evaluation is not possible. Instead, we have to manipulate the expression or apply special techniques to progress. Indeterminate forms indicate that further work is needed to understand the behavior of the function at that point. For instance, in the given problem, we initially encounter the form 0鈰(-鈭). By creatively rearranging terms or using L'H么pital's Rule, we can convert it to a more easily solvable expression. Remember, recognizing and resolving indeterminate forms is crucial in calculus problem solving.
An Introduction to Limits in Calculus
In calculus, limits help us understand the behavior of functions as they approach a certain point or infinity. A limit can describe how a function behaves as the input approaches a specific value. This is central in defining concepts like continuity, derivatives, and integrals.
For example, consider the expression \(\lim_{x \rightarrow 0^{+}}(4x^{2})(\ln x)\). As \(x\) approaches 0 from the positive side, the function's elements \(4x^2\) and \(\ln x\) behave differently. The square term typically tends toward zero, while the natural logarithm approaches negative infinity. This creates an indeterminate form needing careful analysis. Using advanced techniques like L'H么pital's Rule allows further evaluation. Such approaches make it possible to pin down a value and better understand the intricate behavior near limits.
For example, consider the expression \(\lim_{x \rightarrow 0^{+}}(4x^{2})(\ln x)\). As \(x\) approaches 0 from the positive side, the function's elements \(4x^2\) and \(\ln x\) behave differently. The square term typically tends toward zero, while the natural logarithm approaches negative infinity. This creates an indeterminate form needing careful analysis. Using advanced techniques like L'H么pital's Rule allows further evaluation. Such approaches make it possible to pin down a value and better understand the intricate behavior near limits.
Calculus Problem Solving Using L'H么pital's Rule
L'H么pital's Rule is a powerful tool for solving limit problems that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Named after the French mathematician Guillaume de l'H么pital, the rule states that under certain conditions, the limit of a function can be found by taking the derivative of the numerator and the denominator separately.
In our exercise, after converting the initial problem to the form \( \frac{-\infty}{\infty} \), L'H么pital's Rule is applied. By differentiating the numerator \(\ln x\) and the denominator \(1/(4x^2)\), and then evaluating the new expression, the rule simplifies the problem considerably. It changes a seemingly complex limit into a straightforward calculation. This results in a much clearer path to finding the solution鈥攗ltimately obtaining the answer to be zero.
In our exercise, after converting the initial problem to the form \( \frac{-\infty}{\infty} \), L'H么pital's Rule is applied. By differentiating the numerator \(\ln x\) and the denominator \(1/(4x^2)\), and then evaluating the new expression, the rule simplifies the problem considerably. It changes a seemingly complex limit into a straightforward calculation. This results in a much clearer path to finding the solution鈥攗ltimately obtaining the answer to be zero.
- L'H么pital's Rule is valid when indeterminate forms are correctly identified and conditions for differentiability are met.
- Always double-check if the problem can be solved by simpler means before applying the rule.