Problem 39 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0^{+}}\left(\frac{1}{\sin x}-\frac{1}{x}\right) $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify Form of Expression

The limit we need to evaluate is \( \lim _{x \rightarrow 0^{+}}\left(\frac{1}{\sin x}-\frac{1}{x}\right) \). As \( x \to 0^+ \), both \( \frac{1}{\sin x} \) and \( \frac{1}{x} \) approach infinity. Thus, the expression is of the indeterminate form \( \frac{\infty - \infty}{1} \).

Step 2: Combine Fractions

To apply l'Hospital's rule, we need a single fraction. Combine the expressions under a common denominator: \( \frac{x - \sin x}{x \sin x} \). As \( x \rightarrow 0^{+} \), this is an indeterminate form \( \frac{0}{0} \).

Step 3: Apply l'Hospital's Rule

Apply l'Hospital's rule: differentiate the numerator and the denominator. The derivative of the numerator \( x - \sin x \) is \( 1 - \cos x \), and the derivative of the denominator \( x \sin x \) is \( \sin x + x \cos x \). Hence, the limit becomes \( \lim _{x \rightarrow 0^+} \frac{1 - \cos x}{\sin x + x \cos x} \).

Step 4: Evaluate the New Limit

Evaluate the new limit \( \lim _{x \rightarrow 0^+} \frac{1 - \cos x}{\sin x + x \cos x} \). As \( x \rightarrow 0^+ \), \( 1 - \cos x \rightarrow 0 \) and \( \sin x + x \cos x \rightarrow 0 \) still forms \( \frac{0}{0} \). Apply l'Hospital's rule again.

Step 5: Differentiate Again

Differentiate the new numerator and denominator: the derivative of \( 1 - \cos x \) is \( \sin x \), and the derivative of \( \sin x + x \cos x \) is \( \cos x + \cos x - x \sin x = 2\cos x - x\sin x \). Thus, the limit becomes \( \lim _{x \rightarrow 0^+} \frac{\sin x}{2\cos x - x\sin x} \).

Step 6: Evaluate Final Limit

Evaluate the limit \( \lim _{x \rightarrow 0^+} \frac{\sin x}{2\cos x - x\sin x} \). As \( x \rightarrow 0^+ \), the numerator approaches \( 0 \) and the denominator approaches \( 2 \), as \( 2\cos x - x\sin x \rightarrow 2 \). The limit thus evaluates to \( \frac{0}{2} = 0 \).

Key Concepts

Limits in Calculus

Limits are a fundamental concept in calculus. They describe the value that a function approaches as the input approaches a given point. Understanding limits helps us analyze the behavior of functions and is essential for defining derivatives and integrals.

Some key aspects of limits include:

• A limit does not need to be reached. It's about getting infinitely close to a certain point.
• If a limit exists, the function approaches a specific value as the input approaches a point.
• Limits help us work with continuous functions and understand their properties.

In our exercise, we were tasked to evaluate the limit\[\lim _{x \rightarrow 0^{+}}\left(\frac{1}{\sin x}-\frac{1}{x}\right).\]As \( x \) approaches 0 from the positive side, both \( \frac{1}{\sin x} \) and \( \frac{1}{x} \) become very large, making the expression indeterminate. This is where l'Hospital's rule comes in handy.

Indeterminate Forms

Indeterminate forms often occur when evaluating limits and result in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms show that straightforward calculation will not directly lead to a useful answer.

In our exercise, we encountered the form \( \frac{\infty - \infty}{1} \), which is indeterminate when seen as separate fractions. By combining these into a single expression \( \frac{x - \sin x}{x \sin x} \), we transformed it into the indeterminate form \( \frac{0}{0} \). This transformation is crucial as it allows us to applyl'Hospital's rule to find a more specific result.

Indeterminate forms signal that further algebraic manipulation or specific limit rules, like l'Hospital's Rule, need to be applied to resolve the expression.

Differentiation

Differentiation is the process of finding the derivative of a function, which gives us the rate at which the function's value changes as the input changes. It's vital for solving limits, particularly when using l'Hospital's rule.

Here's how differentiation was used in our exercise:

• The derivative of \( x - \sin x \) is \( 1 - \cos x \), found through basic derivative rules.
• The derivative of \( x \sin x \) utilizes both product and trigonometric differentiation rules, resulting in \( \sin x + x \cos x \).
• By continuing to differentiate during subsequent steps, we obtained \( \sin x \) for the numerator and \( 2\cos x - x\sin x \) for the denominator.

These derivatives allow us to express the limit in a form where both the numerator and denominator's behavior at \( x = 0^+ \) can be critically analyzed, and the limit simplification becomes straightforward. Finally, evaluating these expressions facilitates finding the final limit as 0.