Problem 28 - Thomas Calculus

Question

L'Hopital's Rule does not help with the limits in Exercises \(27-30 .\) 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: Analyze the Limit Form

Evaluate the form of the given limit: \( \lim_{x \to 0^{+}} \frac{\sqrt{x}}{\sqrt{\sin x}} \). Both \( \sqrt{x} \) and \( \sqrt{\sin x} \) approach zero as \( x \to 0^{+} \), resulting in an indeterminate form \( \frac{0}{0} \). This indicates that L'Hopital's Rule could be applied, but we've been told it's not helpful for this problem.

Step 2: Simplify the Expression

Use the Taylor series expansion for \( \sin x \) near \( x=0 \), which is \( \sin x \approx x - \frac{x^3}{6} + \cdots \). Thus, for small \( x \), \( \sqrt{\sin x} \approx \sqrt{x} \sqrt{1 - \frac{x^2}{6} + \cdots} \). Factor out \( \sqrt{x} \) from both the numerator and the denominator.

Step 3: Apply the Square Root Identity

Rewrite \( \sqrt{\sin x} \approx \sqrt{x} \left(1 - \frac{x^2}{12} + \cdots \right) \) using the approximation that \( \sqrt{1-y} \approx 1 - \frac{y}{2} \) for small \( y \). Thus, divide both the numerator and denominator by \( \sqrt{x} \): \( \lim_{x \to 0^{+}} \frac{1}{\sqrt{1 - \frac{x^2}{6}}} \).

Step 4: Evaluate the Simplified Limit

As \( x \to 0^{+} \), the term \( \frac{x^2}{6} \) in \( \sqrt{1 - \frac{x^2}{6}} \to 0 \). So, the expression \( \frac{1}{\sqrt{1 - \frac{x^2}{6}}} \) approaches \( \frac{1}{1} = 1 \). Therefore, the limit is \( 1 \).

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a handy tool in calculus for evaluating limits, particularly those that initially result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It states that if the limit of functions \( f(x) \) and \( g(x) \) as \( x \to c \) results in an indeterminate form, we can differentiate the numerator and the denominator. This turns the limit into a potentially solvable form:

• If \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided this limit exists.

However, some limits continuously cycle through indeterminate forms even after applying L'Hopital's Rule, as seen in the given problem. This is why it's crucial to identify when other approaches, like Taylor Series, might offer better solutions.

Limits

In calculus, a limit determines the behavior of a function as the input approaches a specific value. It helps us understand how a function behaves near certain points and is foundational for concepts like continuity and derivatives.

• When evaluating \( \lim_{x \to 0^{+}} \frac{\sqrt{x}}{\sqrt{\sin x}} \), both numerator and denominator approach zero as \( x \) approaches zero from the positive side, which leads us to the indeterminate form \( \frac{0}{0} \).
• Though L'Hopital’s Rule often aids in solving such limits, it is not always the easiest or the only method, as demonstrated here.

Understanding and recognizing limits and their forms allow us to apply other methods effectively and determine the behavior of more complex functions.

Taylor Series

The Taylor Series is a powerful tool in calculus for approximating functions around a point. It expresses functions as infinite sums of terms calculated from the values of its derivatives at a single point.

• For example, the Taylor Series expansion of \( \sin x \) around \( x=0 \) is \( \sin x \approx x - \frac{x^3}{6} + \cdots \).

This approximation is particularly useful in solving limits when functions behave differently around the point of interest, providing transparency to otherwise complex expressions.Using the Taylor Series, we approximate \( \sqrt{\sin x} \) by factoring out \( \sqrt{x} \) and using the expansion for small values of \( x \), simplifying our expression into a form where the limit is simpler to compute.

Indeterminate Forms

Indeterminate forms arise in calculus when standard algebraic methods fail to directly evaluate a limit. Common types of indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \times \infty \)

These forms need further investigation using techniques such as algebraic manipulation, L'Hopital's Rule, or series approximation to find meaningful limits.In our problem, the limit of \( \lim_{x \to 0^{+}} \frac{\sqrt{x}}{\sqrt{\sin x}} \) results in the indeterminate form \( \frac{0}{0} \), suggesting that straightforward substitution doesn’t work. Recognizing this indeterminate form leads us to use other methods, like the Taylor Series, to find the limit effectively and accurately to be \( 1 \).