Problem 23 - Calculus

Question

Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{x \rightarrow 0} \frac{\sin x}{\sin x / 2} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Understand the Function

We need to find the limit \[ \lim_{x \to 0} \frac{\sin x}{\sin \frac{x}{2}} \]. It is important to note that both the numerator \( \sin x \) and the denominator \( \sin \frac{x}{2} \) will approach 0 as \( x \to 0 \). This suggests a \( \frac{0}{0} \) indeterminate form, so we will use L'Hôpital's Rule or trigonometric identities to solve the limit.

Step 2: Apply Trigonometric Identity

We know from trigonometric identities that for small angles, \( \sin x \approx x \). Applying this to both the numerator and the denominator gives us:\[ \lim_{x \to 0} \frac{\sin x}{\sin \frac{x}{2}} \approx \lim_{x \to 0} \frac{x}{\frac{x}{2}} = \lim_{x \to 0} 2 = 2\].

Step 3: Validation Using L'Hôpital's Rule

Although we have already determined the limit, let's confirm it using L'Hôpital's Rule. L'Hôpital's Rule states that if we have an indeterminate form like \( \frac{0}{0} \), we can take the derivative of the numerator and the denominator and then recalibrate the limit.We find the derivatives:- Derivative of \( \sin x \) is \( \cos x \).- Derivative of \( \sin \frac{x}{2} \) with respect to \( x \) is \( \frac{1}{2} \cos \frac{x}{2} \).Applying L'Hôpital's Rule:\[ \lim_{x \to 0} \frac{\cos x}{\frac{1}{2} \cos \frac{x}{2}} = \lim_{x \to 0} \frac{2 \cos x}{\cos \frac{x}{2}} \].As \( x \to 0 \), both \( \cos x \) and \( \cos \frac{x}{2} \) approach 1. Therefore, the limit becomes:\[ \frac{2(1)}{1} = 2 \]. This matches our result from Step 2.

Key Concepts

Trigonometric Identities

Trigonometric identities are crucial tools in calculus, especially when dealing with limits involving trigonometric functions. A well-known identity is that for small angles measured in radians, we have the approximation: \( \sin x \approx x \). This is particularly useful in the limit problem \( \lim_{x \to 0} \frac{\sin x}{\sin \frac{x}{2}} \). As \( x \) approaches zero, both \( \sin x \) and \( \sin \frac{x}{2} \) tend towards zero, making direct substitution impossible due to the \( \frac{0}{0} \) indeterminate form.
To resolve this, by recognizing that for small angles, \( \sin x \approx x \), we can approximate \( \sin \frac{x}{2} \approx \frac{x}{2} \). Substituting these approximations transforms the original limit into:

• \( \frac{x}{\frac{x}{2}} = 2 \)

Hence, by using small angle approximations, trigonometric identities simplify complex limits into easily calculable forms.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful method for evaluating the limits of indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It states that if a limit takes the form \( \frac{0}{0} \), the limit of the ratio of the derivatives of the functions can be taken instead. Consider the original problem:

\[ \lim_{x \to 0} \frac{\sin x}{\sin \frac{x}{2}} \]
This is a \( \frac{0}{0} \) indeterminate form. According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \sin \frac{x}{2} \) with respect to \( x \) is \( \frac{1}{2} \cos \frac{x}{2} \).

Application of L'Hôpital's Rule yields:

\[ \lim_{x \to 0} \frac{\cos x}{\frac{1}{2} \cos \frac{x}{2}} = \lim_{x \to 0} \frac{2 \cos x}{\cos \frac{x}{2}} \]
As \( x \to 0 \), \( \cos x \) and \( \cos \frac{x}{2} \) both approach 1. Thus, the result is:

• \( \frac{2 \cdot 1}{1} = 2 \)

L'Hôpital's Rule confirms the limit calculation obtained using trigonometric identities.

Indeterminate Forms

The term "indeterminate form" refers to an expression that does not initially have a conclusive value when certain functions approach particular points. The form \( \frac{0}{0} \) is one such case. These forms frequently arise in calculus, especially when finding limits involving trigonometric or exponential expressions.

In our limit problem, \( \lim_{x \to 0} \frac{\sin x}{\sin \frac{x}{2}} \), both the numerator \( \sin x \) and the denominator \( \sin \frac{x}{2} \) approach 0 as \( x \to 0 \), causing a \( \frac{0}{0} \) indeterminate form.

• This indeterminate nature prompts the use of methods like L'Hôpital's Rule or trigonometric approximations to evaluate the limit successfully.
• Through insightful methods like re-evaluating derivatives or using approximations, indeterminate forms can be resolved into defined values.

Understanding and recognizing indeterminate forms is essential for advancing in calculus and tackling complex limit problems efficiently.

Small Angle Approximation

Small angle approximation is a practical concept mainly used when dealing with angles close to zero measured in radians. For angles near zero, trigonometric functions have simplified representations that closely approximate the angle itself.

• \( \sin x \approx x \)
• \( \cos x \approx 1 \)
• \( \tan x \approx x \)

These approximations enable easier computation of limits that involve small angles, especially when they result in indeterminate forms.
Let's revisit the limit we confronted:

\[ \lim_{x \to 0} \frac{\sin x}{\sin \frac{x}{2}} \]
Applying small angle approximation, we get:

• \( \sin x \approx x \)
• \( \sin \frac{x}{2} \approx \frac{x}{2} \)

So the limit simplifies to:

• \( \frac{x}{\frac{x}{2}} = 2 \)

This concept is not only important in solving limits but also in physics and engineering where precise trigonometric calculations for small angles are required.