Problem 17
Question
$$ \lim _{x \rightarrow 0^{+}} \frac{x^{2}}{\sin x-x} $$
Step-by-Step Solution
Verified Answer
The limit approaches \(+\infty\) as \( x \rightarrow 0^+ \).
1Step 1: Understand Limits
We need to evaluate \( \lim_{x \to 0^+} \frac{x^2}{\sin x - x} \). This is the right-hand limit as \( x \) approaches 0. Observe that as \( x \to 0 \), both the numerator \( x^2 \) and the expression \( \sin x - x \) in the denominator approaches 0. This is an indeterminate form of \( \frac{0}{0} \), and it suggests we use L'Hôpital's Rule.
2Step 2: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, L'Hôpital's Rule applies: \( \lim_{x \to 0^+} \frac{f(x)}{g(x)} = \lim_{x \to 0^+} \frac{f'(x)}{g'(x)} \) if the limit exists. Here, \( f(x) = x^2 \) and \( g(x) = \sin x - x \).
3Step 3: Differentiate the Numerator and Denominator
Differentiate \( f(x) = x^2 \), giving \( f'(x) = 2x \). Differentiate \( g(x) = \sin x - x \), giving \( g'(x) = \cos x - 1 \).
4Step 4: Evaluate the Limit of Derivatives
Now consider \( \lim_{x \to 0^+} \frac{f'(x)}{g'(x)} = \lim_{x \to 0^+} \frac{2x}{\cos x - 1} \). Again, as \( x \to 0 \), both the numerator and denominator approach 0 \((\frac{0}{0})\). Apply L'Hôpital's Rule again.
5Step 5: Apply L'Hôpital's Rule Again
Differentiate \( 2x \) to get \( 2 \) and \( \cos x - 1 \) to get \( -\sin x \). Evaluate the new limit: \( \lim_{x \to 0^+} \frac{2}{-\sin x} = \lim_{x \to 0^+} \frac{-2}{\sin x} \).
6Step 6: Find the Final Limit
As \( x \to 0^+ \), \( \sin x \to 0 \), making the limit approach \( -\infty \). This indicates that the original limit approaches \( +\infty \) since as we approach from the right \( \lim_{x \to 0^+} \frac{-2}{\sin x} \rightarrow -\infty \). Therefore, the given limit evaluates to \( +\infty \).
Key Concepts
L'Hôpital's RuleIndeterminate FormsCalculus Problem-solvingDifferentiation
L'Hôpital's Rule
L'Hôpital's Rule is an important technique in calculus used to find limits of indeterminate forms. When directly substituting a point into a limit yields an undefined form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule can help solve it. The rule states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) given that the limit on the right exists. This means we differentiate the numerator and the denominator separately, then find the limit of the new fraction. It's important to note that L'Hôpital's Rule can be applied repeatedly if the resulting form is still indeterminate.
Indeterminate Forms
Limits often lead to what are called indeterminate forms. These forms such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), and others, do not readily provide a specific result or direction. This is why special techniques like L'Hôpital's Rule are employed to resolve them. When you encounter an indeterminate form, it signals the need for further analysis to compute the limit properly. The purpose here is to manipulate or transform the existing problem into one with a determinate form which can be evaluated directly.
Calculus Problem-solving
Solving calculus problems often involves a strategic approach for breaking down complex expressions. Here is a general approach:
- Understand the Problem: Start by clearly defining what the problem is asking. Check for any possible indeterminate forms.
- Choose the Right Method: Selection of the correct technique such as L'Hôpital's Rule, algebraic manipulation, or factoring is crucial.
- Step-by-step Approach: Solve the problem step by step without skipping any transformations or calculations.
- Verify Results: Always plug back your solution or use an alternative method to verify the results.
Differentiation
Differentiation is a cornerstone concept in calculus that refers to finding the derivative of a function. Derivatives measure how a function changes as its input changes, and they are represented by \( f'(x) \) or \( \frac{dy}{dx} \). In practical terms, the derivative gives us the rate of change of one variable with respect to another. Differentiation is used in L'Hôpital's Rule to transform an indeterminate form into one that can be evaluated. This is done by finding the derivatives of the given functions in the numerator and the denominator. Each differentiation step should be done carefully to ensure the correct transformation from the original limit expression.
Other exercises in this chapter
Problem 16
$$ \lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x^{2} \sin x} $$
View solution Problem 16
Evaluate each improper integral or show that it diverges. \(\int_{4}^{\infty} \frac{d x}{(\pi-x)^{2 / 3}}\)
View solution Problem 17
Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^{2}+9}} d x\)
View solution Problem 17
Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow(\pi / 2)^{-}}(5 \cos x)^{\tan x}$$
View solution