Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{\sin x-x}{x^{2}} $$

Short answer

The limit as \( x \to 0 \) is 0.

Step-by-step solution

Identifying the Indeterminate Form

Substitute \( x = 0 \) into the limit expression to determine if it presents an indeterminate form. The function becomes \( \frac{\sin(0)-0}{0^2} = \frac{0}{0} \), an indeterminate form.

Applying L'Hôpital's Rule

Since the limit is of the form \( \frac{0}{0} \), we can apply L'Hôpital's rule, which involves taking the derivative of the numerator and the denominator separately. The derivative of \( \sin x - x \) is \( \cos x - 1 \), and the derivative of \( x^2 \) is \( 2x \).

Evaluating New Limit Expression

Rewrite the limit using the derivatives: \( \lim_{x \to 0} \frac{\cos x - 1}{2x} \). Substituting \( x = 0 \), we get \( \frac{\cos(0) - 1}{0} = \frac{0}{0} \), which is still indeterminate.

Applying L'Hôpital's Rule Again

Apply L'Hôpital's rule a second time: differentiate the numerator and denominator again. The derivative of \( \cos x - 1 \) is \( -\sin x \), and the derivative of \( 2x \) is \( 2 \).

Evaluating Final Limit Expression

Rewrite the limit after the second application of L'Hôpital's rule: \( \lim_{x \to 0} \frac{-\sin x}{2} \). Substituting \( x = 0 \), we find \( \frac{-\sin(0)}{2} = \frac{0}{2} = 0 \).

Conclusion

The limit as \( x \) approaches 0 is \( 0 \). Hence, \( \lim_{x \rightarrow 0} \frac{\sin x-x}{x^{2}} = 0 \).

Key concepts

Indeterminate Forms

Indeterminate forms are expressions where traditional arithmetic rules do not provide a clear result. They often arise in calculus when evaluating limits. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), and several others. In our exercise, substituting \( x = 0 \) into the limit expression \( \frac{\sin(x) - x}{x^2} \) results in \( \frac{0}{0} \), which is an indeterminate form. This happens because both the numerator and denominator approach zero as \( x \) approaches zero, leaving the fraction's value uncertain.
The beauty of indeterminate forms is that they allow us to employ specialized techniques, such as L'Hôpital's Rule, to evaluate limits that initially seem unsolvable.

Derivatives

Derivatives are fundamental to calculus and are used to determine rates of change. In the context of L'Hôpital's Rule, they allow us to simplify complex limits, especially when dealing with indeterminate forms. For the given function \( \frac{\sin(x) - x}{x^2} \), we take the derivative of the numerator \( \sin x - x \), which is \( \cos x - 1 \), and the derivative of the denominator \( x^2 \), which is \( 2x \).
This process transforms our original limit into a potentially more solvable form by analyzing how each part changes as \( x \) approaches the point of interest. The transformation from the original function to its derivatives gives us a clearer picture that can help resolve the indeterminate form. In this exercise, a second derivative is also needed, where \( -\sin x \) and \( 2 \) become the new numerator and denominator, respectively.
Understanding derivatives in this manner is crucial for effectively applying L'Hôpital's Rule.

Calculus Limits

Calculus limits are the backbone of calculus, allowing us to understand the behavior of functions as inputs approach certain values. When encountering a limit that results in an indeterminate form, as seen with \( \lim_{x \rightarrow 0} \frac{\sin x - x}{x^2} \), our goal is to explore the function's behavior around \( x = 0 \).
L'Hôpital's Rule serves as a powerful tool to tackle such limits by allowing us to look beyond the immediate indeterminate form. Once derivatives are appropriately applied and simplify the limit expression, evaluating the limit becomes straightforward. For example, after using the derivatives twice in our exercise, the expression \( \lim_{x \to 0} \frac{-\sin x}{2} \) doesn't pose an indeterminate form anymore.
Mastering how calculus limits work, combined with techniques like L'Hôpital's Rule, opens up limitless possibilities for solving complex problems in calculus.