Problem 55 - Calculus: Early Transcendentals

Question

Evaluate limit. $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-1}{\sqrt{\sin x}-1}$$

Step-by-Step Solution

Answer

Answer: The limit of the given function as x approaches π/2 is 2.

Step 1: Identify indeterminate form

First, let's plug in x = π/2 in the given function to see if it is indeterminate: $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-1}{\sqrt{\sin x}-1}=\frac{\sin (\pi/2)-1}{\sqrt{\sin (\pi/2)}-1}=\frac{0}{0}$$ Since we have the indeterminate form 0/0, we can use L'Hôpital's Rule or algebraic manipulations to simplify the function.

Step 2: Simplify the function using algebraic manipulation

To remove the indeterminate form, we'll multiply the numerator and denominator of the expression by the conjugate of the denominator: $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-1}{\sqrt{\sin x}-1} \cdot \frac{\sqrt{\sin x}+1}{\sqrt{\sin x}+1}$$ Expanding the expression, we get: $$\lim _{x \rightarrow \pi / 2} \frac{(\sin x -1)(\sqrt{\sin x} +1)}{(\sqrt{\sin x}-1)(\sqrt{\sin x}+1)}$$ Now, we simplify the denominator by using the difference of squares: $$\lim _{x \rightarrow \pi / 2} \frac{(\sin x-1)(\sqrt{\sin x}+1)}{\sin x -1}$$ Now, we can cancel out the (sin x - 1) term from both numerator and denominator: $$\lim _{x \rightarrow \pi / 2} (\sqrt{\sin x}+1)$$

Step 3: Evaluate the limit

Now that we have simplified the function, let's evaluate the limit as x approaches π/2: $$\lim _{x \rightarrow \pi / 2} (\sqrt{\sin x}+1)=\sqrt{\sin(\pi/2)}+1=1 + 1=2$$ Thus, the limit of the given function as x approaches π/2 is 2.

Key Concepts

Understanding Limits

Limits are a fundamental concept in calculus. They help us understand how a function behaves as the input approaches a particular point. In the exercise, we aim to find how the function \( \frac{\sin x - 1}{\sqrt{\sin x} - 1} \) behaves as \( x \) approaches \( \frac{\pi}{2} \).

• Think of limits as the value a function gets closer to as the input approaches a specific number.
• We write the limit of a function \( f(x) \) as \( x \) approaches \( a \) using the notation \( \lim_{x \to a} f(x) \).
• Understanding limits is key for evaluating expressions that seem undefined by direct substitution, like \( \frac{0}{0} \).

When evaluating the exercise, plugging in \( x = \frac{\pi}{2} \) resulted in an expression of \( \frac{0}{0} \), which means further analysis is essential to find the actual limit.

Dealing with Indeterminate Forms

Indeterminate forms arise in calculus when substituting a value into a function leads to expressions like \( \frac{0}{0} \) or \( \infty - \infty \). These forms don't immediately reveal the limit's direction or value, prompting the use of advanced methods to resolve them.

• An indeterminate form means we cannot determine the limit directly and need another approach.
• Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and analogous combinations.
• We utilize tools like L'Hôpital's Rule or algebraic manipulation to remove indeterminacy.

In the problem, we started with \( \frac{0}{0} \), requiring us to employ strategies like multiplying by the conjugate to simplify and find a clear expression for evaluating the limit.

Algebraic Manipulation in Limits

Algebraic manipulation involves rearranging expressions to simplify or clarify problems. It is a valuable skill when dealing with indeterminate forms, allowing us to remove problematic parts of expressions.

• Algebraic techniques can include factoring, expanding, or rationalizing expressions.
• In the problem, the expression was manipulated by multiplying by the conjugate \( \sqrt{\sin x}+1 \).
• This technique transformed the denominator's expression \( (\sqrt{\sin x}-1)(\sqrt{\sin x}+1) \) into \( \sin x - 1 \), allowing cancellation with the numerator.

Through algebraic manipulation, we simplified the limit to \( \lim_{x \rightarrow \pi / 2} (\sqrt{\sin x} + 1) \), making it straightforward to evaluate, resulting in the limit being 2.