Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{-}} \frac{3 \sin x}{\sqrt{-x}} $$

Short answer

The limit is 0.

Step-by-step solution

Evaluate the Limit at the Given Point

First, substitute \( x = 0^{-} \) into the expression \( \frac{3 \sin x}{\sqrt{-x}} \) to see if it is an indeterminate form.Substituting gives:\[\frac{3 \sin(0^{-})}{\sqrt{-(0^{-})}} = \frac{0}{0}\]Since \( \frac{0}{0} \) is an indeterminate form, we can apply l'Hôpital's Rule.

Apply l'Hôpital's Rule

According to l'Hôpital's Rule, if you have a limit of the form \( \frac{0}{0} \), you can take the derivative of the numerator and the denominator to evaluate the limit.We start by finding the derivative of the numerator:\[\frac{d}{dx}[3 \sin x] = 3 \cos x\]Now, find the derivative of the denominator:\[\frac{d}{dx}[\sqrt{-x}] = \frac{d}{dx}[(-x)^{1/2}] = \frac{-1}{2\sqrt{-x}}\]Now the limit becomes:\[\lim_{x \rightarrow 0^{-}} \frac{3 \cos x}{-\frac{1}{2\sqrt{-x}}}\]

Simplify the New Expression

Simplify the expression obtained after applying l'Hôpital's Rule:Rewriting the denominator gives:\[\lim_{x \rightarrow 0^{-}} \frac{3 \cos x}{-\frac{1}{2\sqrt{-x}}} = \lim_{x \rightarrow 0^{-}} -6 \cos x \sqrt{-x}\]

Evaluate the Simplified Limit

Now, evaluate the limit as \( x \rightarrow 0^{-} \):As \( x \) approaches \( 0^{-} \), \( \cos x \) approaches \( \cos 0 = 1 \), and \( \sqrt{-x} \) approaches \( 0 \).Thus, the simplified expression \(-6 \cos x \sqrt{-x}\) approaches \( 0 \).Therefore, the limit is:\[0\]

Key concepts

Indeterminate Forms

In calculus, indeterminate forms arise when evaluating limits that produce expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms are considered 'indeterminate' because they do not directly provide enough information about the actual behavior of the function in those limits. This makes it challenging to predict what the limit will resolve to without further analysis.

To deal with these forms, mathematicians employ various techniques. One powerful tool is l'Hôpital's Rule, which assists in resolving limits by utilizing derivatives. Recognizing an indeterminate form is crucial. It indicates that the limit needs additional work to determine a meaningful result, steering you towards methods like l'Hôpital's Rule for resolution.

The original exercise gives an example of an indeterminate form \( \frac{0}{0} \), which allows us to apply further calculus techniques to evaluate the limit.

Derivatives

Derivatives lie at the heart of calculus and are essential when utilizing l'Hôpital's Rule. A derivative represents the rate of change of a function concerning a variable. When applying l'Hôpital's Rule, we compute the derivatives of the numerator and the denominator of the limit function separately.

In the given exercise, we first determined the derivative of the numerator, \(3 \sin x\), which results in \(3 \cos x\). For the denominator, \(\sqrt{-x}\), we use the power rule for derivatives, yielding \(-\frac{1}{2\sqrt{-x}}\).

These derivatives replace the original functions in the limit, allowing a new limit expression to be formed, which ideally no longer presents an indeterminate form. This is the core mechanism behind l'Hôpital's Rule, making derivatives a crucial step in resolving complex limits.

Limit Evaluation

Limit evaluation is a fundamental process in calculus for determining the behavior of a function as the input approaches a particular value. Properly evaluating limits often requires identifying potential indeterminate forms, especially as variables approach zero or infinity.

In this exercise, after applying l'Hôpital's Rule, the new expression for the limit was \(-6 \cos x \sqrt{-x}\). Evaluating this expression as \(x\) approaches \(0^-\) involves understanding the behavior of \(\cos x\) and \(\sqrt{-x}\).

• \(\cos x\) is known to approach \(1\) as \(x\) gets closer to \(0\).
• The term \(\sqrt{-x}\) approaches \(0\) because \(x\) itself approaches \(0^-\).

The product thus simplifies neatly to \(0\), giving us the final evaluated limit. This demonstrates the clear utility of methodical limit evaluation in addressing complex calculus problems.

Trigonometric Limits

Trigonometric functions, like \(\sin\) and \(\cos\), frequently appear in limit problems. Understanding their behavior is essential for evaluating limits that involve these functions. In our exercise, the function \(3 \sin x\) is present in the original limit setup, leading to the initial indeterminate form.

When evaluating trigonometric limits, especially around zero, we rely on fundamental identities and limits, such as the well-known results:

• \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)
• \(\cos 0 = 1\)

These simplifications are crucial for determining how the components of the trigonometric function behave as they approach their limits.

In our problem, \(\cos x\) approaches \(1\) as \(x\) nears zero, allowing the resolution of the original limit expression effectively. This illustrates the importance of familiarity with trigonometric limits, which are pivotal in solving a variety of calculus problems involving these functions.