Problem 56 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} \frac{1-\cos x}{x} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Indeterminate Form

The expression \( \lim _{x \rightarrow 0} \frac{1-\cos x}{x} \) initially gives \( \frac{0}{0} \) when substituting \( x = 0 \). This indicates an indeterminate form and suggests the need for further analysis or application of techniques to resolve it.

Step 2: Use Trigonometric Identity

Recall the trigonometric identity \( 1 - \cos x = 2 \sin^2(\frac{x}{2}) \). Substitute this identity into the original limit expression: \[\lim _{x \rightarrow 0} \frac{1-\cos x}{x} = \lim _{x \rightarrow 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x}\]

Step 3: Simplify the Expression

Rewrite the expression using the substitution \( u = \frac{x}{2} \) so that as \( x \to 0 \), \( u \to 0 \) as well. This gives us:\[\lim _{u \rightarrow 0} \frac{2 \sin^2(u)}{2u}\]which simplifies to:\[\lim _{u \rightarrow 0} \frac{\sin^2(u)}{u}\]

Step 4: Use Known Trigonometric Limit

Recall the fundamental trigonometric limit: \( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \). Therefore,\[\lim_{u \to 0} \frac{\sin^2(u)}{u} = \lim_{u \to 0} \sin(u) \cdot \frac{\sin(u)}{u} = \lim_{u \to 0} \sin(u) \cdot 1 = \lim_{u \to 0} \sin(u)\]

Step 5: Evaluate the Final Limit

As \( u \to 0 \), \( \sin(u) \to 0 \). Thus, the overall limit is:\[\lim _{x \rightarrow 0} \frac{1-\cos x}{x} = 0\]

Key Concepts

l'Hôpital's Rule

In calculus, l'Hôpital's rule is a technique used to find limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you plug a number into a limit expression and end up with one of these forms, l'Hôpital's rule can often help. Instead of working with the original limit expression directly, you replace it with \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), where \( f(x) \) and \( g(x) \) are functions such that their limit as \( x \to c \) gives an indeterminate form. **It's crucial that both \( f(x) \) and \( g(x) \) are differentiable.**

**When to Use l'Hôpital's Rule:**

• Verify that the limit produces an indeterminate form (e.g., \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)) when the variable approaches a particular value.
• Differentiable functions: Both the numerator and denominator must be differentiable at the point of interest.

However, l'Hôpital's rule isn't always the best or only method to find a limit. In some cases, such as with trigonometric identities or algebraic manipulations, simpler methods might be more efficient. This was the case in our problem, where by using a trigonometric identity, we avoided applying l'Hôpital's rule entirely. This highlights the importance of knowing various mathematical techniques.

Trigonometric Identities

Trigonometric identities are powerful tools in calculus and trigonometry that simplify complex expressions and make solving problems like limits more straightforward. A trigonometric identity is a well-known formula that relates the trigonometric functions to one another. In the solved exercise, we used the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \). This identity simplifies expressions involving the cosine function.

**Key Trigonometric Identity Used:**

• When evaluating the limit \( \lim _{x \to 0} \frac{1-\cos x}{x} \), substituting \( 1 - \cos x \) with \( 2 \sin^2\left(\frac{x}{2}\right) \) helps in simplifying the expression significantly.
• This transformation changes a difficult indeterminate form into an expression that can be more easily tackled using basic trigonometric limits.

Understanding and using trigonometric identities can often provide a more straightforward way to solve calculus problems, especially limits, by reducing the complexity of the expression involved. Instead of more complex calculus techniques, simple algebraic replacements can lead to a solution.

Indeterminate Forms

Indeterminate forms occur frequently when evaluating limits in calculus. These are specific forms which, on first glance, do not provide enough information to determine a limit. The two most common indeterminate forms are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), neither of which has a definitive value, and require additional steps to resolve.

**Understanding Indeterminate Forms:**

• Indeterminate forms such as \( \frac{0}{0} \) suggest that the limiting process has output that approaches zero from both the numerator and the denominator, providing no clear result directly.
• Identifying an indeterminate form is often the first step in problem-solving, prompting further analysis, transformation, or the application of rules like l'Hôpital’s rule.

Encountering an indeterminate form tells us that we need to delve deeper. This might mean employing algebraic manipulation, applying trigonometric identities, or leveraging calculus tricks like differentiation. In our example, recognizing that \( \frac{1-\cos x}{x} \) was of the form \( \frac{0}{0} \) prompted further transformation using trigonometric identities, ultimately leading us to an understandable and solvable form.