Problem 68 - Thomas Calculus in SI Units

Question

L'Hópital's Rule does not help with the limits. Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{\sin x}}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Assess Direct Substitution

Start by substituting 0 directly into the limit expression. As \( x \rightarrow 0^{+} \), both \( \sqrt{x} \) and \( \sqrt{\sin x} \) approach 0. This creates an indeterminate form of \( \frac{0}{0} \), indicating that direct substitution is insufficient, and we need another approach.

Step 2: Utilize Expansion of Functions

Expand \( \sin x \) using the series expansion: \( \sin x \approx x - \frac{x^3}{6} + \cdots \) as \( x \rightarrow 0 \). Therefore, we can approximate \( \sqrt{\sin x} \) by \( \sqrt{x} \) at \( x \rightarrow 0^{+} \). Simplifying, the expression becomes: \[ \frac{\sqrt{x}}{\sqrt{\sin x}} \approx \frac{\sqrt{x}}{\sqrt{x}} = 1 \] when you exclude the higher-order terms.

Step 3: Simplify the Limit Expression

We have already approximated \( \sin x \approx x \) for small values of \( x \). Thus, the expression \( \frac{\sqrt{x}}{\sqrt{\sin x}} \) simplifies directly to \( \frac{\sqrt{x}}{\sqrt{x}} = 1 \) for \( x > 0 \) as we approach the limit.

Step 4: Evaluate the Limit

Since the simplification results in the value 1, we conclude that: \[ \lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{\sin x}} = 1 \]. Hence, the approach using the approximation of \( \sin x \) for small \( x \) confirms the value without further cycles using L'Hôpital's Rule.

Key Concepts

Limits

Understanding limits is crucial, especially when dealing with expressions that change as they approach a particular value. Limits essentially capture the behavior of a function as the input reaches a certain point, even if it never quite gets there. In this exercise, we're examining the limit as \( x \) approaches 0 from the positive side \( (0^+) \). Often, direct computation of limits can be challenging due to the irregular behavior of functions at specific points. Thus, having a solid grasp of limits helps us transform these problems into more approachable forms.

Direct Substitution

Direct substitution is an initial step when evaluating limits. You simply replace the variable with the approaching value. If this method works, it's the quickest way to find a limit. However, challenges arise when substitution results in undefined forms. In this exercise, substituting \( x = 0 \) into \( \sqrt{x} / \sqrt{\sin x} \) leads to \( 0/0 \), an indeterminate form, making direct substitution inadequate. Detecting this early guides us toward alternative techniques, such as series expansions or L'Hôpital's Rule, for further simplification.

Series Expansion

Series expansion is a powerful mathematical tool for simplifying complex expressions near specific points. With series expansion, we rewrite functions in terms of powers of their variables. For example, \( \sin x \) around zero can be expanded as \( \sin x \approx x - \frac{x^3}{6} + \cdots \). This expansion simplifies our limit problem, as it reveals that \( \sin x \) behaves very similarly to \( x \) for small values of \( x \). Consequently, \( \sqrt{\sin x} \) approximates \( \sqrt{x} \), allowing for further simplification of the limit expression.

Indeterminate Forms

Indeterminate forms like \( \frac{0}{0} \) arise when evaluating limits through direct substitution results in expressions that don't immediately yield meaningful values. These forms signal that more sophisticated approaches are required. L'Hôpital's Rule is a common method to resolve these forms, but isn't always effective, as seen in the given exercise. Recognizing indeterminate forms is instrumental, as it prompts us to employ other techniques, such as series expansion, to evaluate the limit accurately, helping us find the genuine behavior of functions as they approach specific points.