Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{+}} \frac{x^{2}}{\sin x-x} $$

Short answer

The limit is 0.

Step-by-step solution

Analyze the Limit Expression

The problem requires us to find the limit \( \lim_{x \to 0^+} \frac{x^2}{\sin x - x} \). First, identify if the expression \( \frac{x^2}{\sin x - x} \) is in an indeterminate form. As \( x \to 0^+ \), both the numerator and the denominator approach 0. Therefore, the expression is in the \( \frac{0}{0} \) form, which allows us to apply l'Hôpital's Rule.

Applying l'Hôpital's Rule

Since the limit is in an indeterminate form \( \frac{0}{0} \), we can use l'Hôpital's Rule, which states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) if the original limit is in an indeterminate form. First, differentiate the numerator \( f(x) = x^2 \), which gives \( f'(x) = 2x \). Now, differentiate the denominator \( g(x) = \sin x - x \), which gives \( g'(x) = \cos x - 1 \).

Calculating the New Limit Using l'Hôpital's Rule

Substitute the derivatives into the limit to apply l'Hôpital's Rule: \( \lim_{x \to 0^+} \frac{2x}{\cos x - 1} \). Evaluate the new limit. As \( x \to 0^+ \), the numerator \( 2x \to 0 \) and the denominator \( \cos x - 1 \to -1 \) since \( \cos(0) = 1 \). Therefore, the limit simplifies to \( 0 \).

Verify the Result

We have applied l'Hôpital's Rule once. Check if the limit still results in an indeterminate form. The result is \( 0 \) because the numerator approaches 0 while the denominator approaches a non-zero number \(-1\). Thus, the limit \( \lim_{x \to 0^+} \frac{2x}{\cos x - 1} = 0 \) is correctly solved.

Key concepts

Indeterminate Forms in Calculus

When dealing with limits, we often encounter situations where direct substitution doesn't provide a clear answer. One common scenario is the indeterminate form. An indeterminate form arises when the result of substituting a value into a limit expression results in a form like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or others such as \( 0^0 \), \( \infty^0 \), or \( 1^\infty \). These forms are termed 'indeterminate' because they do not directly indicate a specific value or behavior.

In this exercise, the original expression \( \frac{x^2}{\sin x - x} \) becomes \( \frac{0}{0} \) as \( x \rightarrow 0^+ \). This means it's neither defined, nor undefined in a straightforward way, making it ripe for application of analytical techniques like l'Hôpital's Rule.

Understanding and identifying indeterminate forms is crucial because they often appear in calculus problems involving limits, especially when derivative evaluation methods like l'Hôpital's Rule are applicable.

Understanding Limits

Limits are a fundamental concept in calculus, serving as the underpinning for derivatives and integrals. A limit describes the behavior of a function as it approaches a particular input. For example, calculating \( \lim_{x \to a} f(x) \) involves understanding how \( f(x) \) behaves as \( x \) gets closer and closer to \( a \).

In cases like \( \lim_{x \to 0^+} \frac{x^2}{\sin x - x} \), direct computation by substitution (\( x = 0 \)) is impossible due to the indeterminate form \( \frac{0}{0} \). Instead, techniques such as simplification, factoring, or applying derivative-related rules like l'Hôpital’s are necessary.

Here, the limit shows how the function behaves just to the right of zero (indicated by \( 0^+ \)). A crucial skill in finding limits is knowing when and how to resolve indeterminate forms so that a meaningful numeric result is achieved.

Effective Calculus Techniques: l'Hôpital's Rule

Calculus offers numerous techniques for solving complex problems, one of the most powerful among them being l'Hôpital's Rule. This rule is a tool used to evaluate limits that at first glance appear indeterminate, typically in the form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

The rule states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the necessary differentiability and continuity conditions are met. This means we differentiate the numerator and the denominator separately, then re-evaluate the limit of the new fraction.

For the example \( \lim_{x \to 0^+} \frac{x^2}{\sin x - x} \), using derivatives transforms it into \( \lim_{x \to 0^+} \frac{2x}{\cos x - 1} \). Evaluating this limit gives \( 0 \), since \( 2x \) approaches 0 and \( \cos x - 1 \) approaches -1 as \( x \rightarrow 0^+ \). Thus, effectively replacing the complex initial expression with a more straightforward one to solve is the beauty of l'Hôpital's Rule.