Problem 8 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0} \frac{x \sin x}{1-\cos x} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Check Indeterminate Form

First, we need to check if the given limit is in an indeterminate form like \( \frac{0}{0} \) before applying l'Hospital's Rule. Evaluate the numerator and denominator at \( x = 0 \):\[ \text{Numerator: } x \sin x \rightarrow 0 \cdot 0 = 0 \] \[ \text{Denominator: } 1 - \cos x \rightarrow 1 - 1 = 0 \] Since both the numerator and denominator approach 0, we have the indeterminate form \( \frac{0}{0} \). Thus, l'Hospital's Rule can be applied.

Step 2: Apply L'Hospital's Rule

L'Hospital's Rule states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is indeterminate as \( \frac{0}{0} \), then \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \] provided the second limit exists. Differentiate the numerator \( f(x) = x \sin x \) to get \( f'(x) = \sin x + x \cos x \). Differentiate the denominator \( g(x) = 1 - \cos x \) to get \( g'(x) = \sin x \).

Step 3: Evaluate the Limit of Derivatives

Now substitute the derivatives back into the limit expression: \[ \lim_{x \to 0} \frac{\sin x + x \cos x}{\sin x} \] Simplify this to: \[ \lim_{x \to 0} \left( 1 + \frac{x \cos x}{\sin x} \right) \] This requires evaluating the limit of \( \frac{x \cos x}{\sin x} \).

Step 4: Simplify and Solve

Returning to \( \frac{x \cos x}{\sin x} \): \[ \lim_{x \to 0} \frac{x \cos x}{\sin x} = \lim_{x \to 0} x \frac{\cos x}{\sin x} = \lim_{x \to 0} x \cdot \cot x \] Since \( x \rightarrow 0 \) and \( \cot x \rightarrow \infty \) near 0, use the approximation \( \cot x \approx \frac{1}{x} \): \[ \lim_{x \to 0} x \cdot \frac{1}{x} = \lim_{x \to 0} 1 = 1 \] Therefore, the original limit evaluates to: \[ \lim_{x \to 0} \left( 1 + \frac{x \cos x}{\sin x} \right) = 1 + 1 = 2 \]

Step 5: Conclusion: Final Answer

By applying l'Hospital's Rule and simplifying, we found that the limit is 2. Therefore, \( \lim _{x \rightarrow 0} \frac{x \sin x}{1-\cos x} = 2 \).

Key Concepts

Indeterminate Forms

Indeterminate forms are expressions in calculus where direct substitution into a limit does not readily provide a defined value. These forms often appear as ratios like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Such forms are "indeterminate" because they do not point to a specific value without further analysis.

When dealing with limits, indeterminate forms signal the need for additional techniques, such as algebraic manipulation, series expansion, or l'Hospital’s Rule. In this exercise, the initial limit \( \lim _{x \rightarrow 0} \frac{x \sin x}{1-\cos x} \) produces an indeterminate \( \frac{0}{0} \) form. Knowing how to recognize these forms is critical in calculus, as it allows you to apply the correct method to find a solution.

Identifying and addressing indeterminate forms help simplify complex limit evaluations, paving the way for techniques that turn seemingly unsolvable problems into manageable ones.

Differentiation

Differentiation is a calculus method used to find the derivative, or the rate of change, of a function. It is integral in understanding how functions behave and change over small intervals.

For l'Hospital's Rule, differentiation is key; it involves differentiating both the numerator and denominator of an indeterminate form. In this exercise, the need arises to differentiate \( f(x) = x \sin x \) and \( g(x) = 1 - \cos x \).

Differentiating gives us the derivatives \( f'(x) = \sin x + x \cos x \) and \( g'(x) = \sin x \). These are then used to simplify and solve the indeterminate limit.

Mastering differentiation will aid in using l'Hospital's Rule effectively and can simplify finding derivatives for other calculus problems, providing essential insights into function behaviors.

Limits in Calculus

Limits in calculus give a way to understand the behavior of a function as it approaches a particular point. This fundamental concept is crucial for grasping more complex topics like continuity, integrals, and derivatives.

In the specific exercise, the notion of a limit helps find the behavior of the expression \( \frac{x \sin x}{1-\cos x} \) as \( x \rightarrow 0 \). By substituting derivatives back into the original expression, the limit simplifies to \( \lim_{x \to 0} \left( 1 + \frac{x \cos x}{\sin x} \right) \), providing a final resolved limit of 2.

Understanding limits empowers you to handle various mathematical scenarios, transitioning smoothly from potential confusion to clarity. This concept serves as a cornerstone for unlocking further studies in calculus and gaining a deeper understanding of how functions behave close to specific values.