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Magazine Chapter 4: Problem 28
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. $$ \lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{\sin x}} $$
Short Answer
Expert verified
The limit is 1.
Step by step solution
01
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.
02
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.
03
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 \( \sqrt{x} \): \( \lim_{x \to 0^{+}} \frac{1}{\sqrt{1 - \frac{x^2}{6}}} \).
04
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 \( 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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鈥檚 Rule often aids in solving such limits, it is not always the easiest or the only method, as demonstrated here.
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 \( \sin x \approx x - \frac{x^3}{6} + \cdots \).
Indeterminate Forms
Indeterminate forms arise in calculus when standard algebraic methods fail to directly evaluate a limit. Common types of indeterminate forms include:
- \( \frac{0}{0} \)
- \( \frac{\infty}{\infty} \)
- \( 0 \times \infty \)
