Problem 39 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0^{+}}\left(\frac{1}{\sin x}-\frac{1}{x}\right) $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Problem

We need to find the limit \( \lim_{x \rightarrow 0^{+}} \left( \frac{1}{\sin x} - \frac{1}{x} \right) \). As \( x \) approaches \( 0 \) from the positive side, both \( \sin x \) and \( x \) approach zero, creating expressions involving division by zero. We look for ways to simplify or transform this to find the limit.

Step 2: Use Known Trigonometric Approximations

Recall that for small \( x \), \( \sin x \approx x \). Use this approximation to understand the behavior of the terms. This simplifies to \( \frac{1}{\sin x} \approx \frac{1}{x} \). Substituting this, the expression becomes \( \frac{1}{x} - \frac{1}{x} \), which is zero, suggesting a limit.

Step 3: Detailed Simplification

Write \( \sin x = x - \frac{x^3}{6} + O(x^5) \) for a more precise approximation. Then, \( \frac{1}{\sin x} = \frac{1}{x - \frac{x^3}{6} + O(x^5)} \). Use the first term in a series: \( \frac{1}{\sin x} \approx \frac{1}{x} \Big(1 + \frac{x^2}{6}\Big) \approx \frac{1}{x} + \frac{x}{6} \). Subtracting \( \frac{1}{x} \), we have \( \frac{x}{6} \).

Step 4: Compute the Limit

Given the simplification \( \frac{1}{\sin x} - \frac{1}{x} = \frac{x}{6} \), the limit as \( x \rightarrow 0^{+} \) is simply the limit of \( \frac{x}{6} \). As \( x \rightarrow 0^{+} \), \( \frac{x}{6} \rightarrow 0 \). Thus, the limit is 0.

Key Concepts

l'Hôpital's Rule

l'Hôpital's Rule is a very useful tool in calculus, specifically designed to tackle indeterminate forms that often appear when evaluating limits. These indeterminate forms typically present as

• Drawing 0/0 (zero over zero)
• Or ∞/∞ (infinity over infinity)

To apply l'Hôpital's Rule, you take the derivative of the numerator and the derivative of the denominator separately. Then, you calculate the limit of this new fraction. If the indeterminate form persists after applying the rule once, you can repeat the process until the limit is resolved.
For example, for \[rac{0}{0} \]indeterminate form like in the example with \[rac{1}{ ext{sin}x} - rac{1}{x},\]the rule can be carefully used. However, understanding the trigonometric behavior without immediately resorting to l'Hôpital’s Rule can make the problem more manageable.

Trigonometric Approximations

Trigonometric approximations, especially for the function \( ext{sin}x \), are extremely helpful when dealing with limits as they approach small values. A fundamental approximation to remember is:

• For very small values of \(x\), \( ext{sin}x \) is approximately equal to \(x\).

This approximation simplifies calculations as it turns complex expression into much simpler forms. It is particularly useful when assessing behavior close to zero, like in our exercise. More precisely, the small-angle approximation tells us that:

• \( ext{sin}x \approx x - \frac{x^3}{6}\)

This is a truncated polynomial series providing an even closer approximation and allowing us to predict the behavior of \(rac{1}{ ext{sin}x}\) around zero.

Series Expansion

Series expansion, particularly Taylor series, can transform difficult functions into manageable expressions. For instance, the sine function can be expanded around zero using the Taylor series, which helps solve limit problems effectively. The sine function \( ext{sin}x\) can be expanded as follows:

• \(x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots\)

This expansion can be shortened to fit the problem at hand, often ignoring higher order terms that have negligible impact on the limit. In our exercise, only the first couple of terms are significant when x is very small, allowing us to approximate:

• \(rac{1}{ ext{sin}x} \approx rac{1}{x} \times \left(1 + \frac{x^2}{6}\right)\)

By employing these principles, you can effectively find limits that initially seem unsolvable by direct substitution.