Problem 51 - A Graphical Approach to Precalculus with Limits

Question

Determine each limit, if it exists. $$\lim _{x \rightarrow 0} \frac{\cos x-1}{3 x}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Recognize the form of the limit

Examine the limit \( \lim _{x \rightarrow 0} \frac{\cos x-1}{3 x} \). Notice that if we substitute \( x = 0 \) directly, the expression becomes \( \frac{\cos(0) - 1}{3 \times 0} = \frac{0}{0} \), which is an indeterminate form. This requires us to use L'Hôpital's Rule or another method to evaluate the limit.

Step 2: Differentiate the numerator and denominator

Since the limit involves an indeterminate form \( \frac{0}{0} \), apply L'Hôpital's Rule, which states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) if the original form is indeterminate. Differentiate the numerator \( \cos x - 1 \) to get \( -\sin x \), and the denominator \( 3x \) to get \( 3 \).

Step 3: Apply L'Hôpital's Rule to find the new limit

Substitute the derivatives back into the limit: \( \lim_{x \to 0} \frac{\cos x - 1}{3x} = \lim_{x \to 0} \frac{-\sin x}{3} \).

Step 4: Evaluate the new limit

Now directly substitute \( x = 0 \) into the new limit \( \lim_{x \to 0} \frac{-\sin x}{3} \) to get \( \frac{-\sin 0}{3} = \frac{0}{3} = 0 \).

Key Concepts

Indeterminate Forms

When working with limits in calculus, you may encounter expressions that are known as "indeterminate forms." These forms need further evaluation because they don't clearly indicate the limit's value. A common indeterminate form is \( \frac{0}{0} \), which happens when both the numerator and the denominator of a fraction approach zero. For example, consider the limit \( \lim_{x \rightarrow 0} \frac{\cos x - 1}{3x} \). Plugging in \( x = 0 \) gives \( \frac{0}{0} \), which doesn't tell us the value of the limit.
This requires additional methods, such as "L'Hôpital's Rule," to resolve. It's crucial to recognize when expressions are indeterminate because simply substituting values would mislead the result. More unusual forms like \( \frac{\infty}{\infty} \), \( 0 \times \infty \), or \( \infty - \infty \) need similar careful analysis.
By transforming these expressions using algebraic manipulations or calculus techniques, we can find well-defined limits.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for solving limits that result in indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule provides a way to simplify these expressions by differentiating the numerator and the denominator and then re-evaluating the limit.
The formal statement of L'Hôpital's Rule is: if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in an indeterminate form, then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), assuming the limit on the right exists.
Let's apply this to \( \lim_{x \rightarrow 0} \frac{\cos x - 1}{3x} \). Both numerator \( \cos x - 1 \) and denominator \( 3x \) become zero when \( x = 0 \), indicating \( \frac{0}{0} \). Differentiating the numerator gives \( - \sin x \) and differentiating the denominator gives \( 3 \). By applying L'Hôpital's Rule, the limit becomes \( \lim_{x \to 0} \frac{-\sin x}{3} \).

Trigonometric Limits

Trigonometric functions often appear in limits, and their behavior near critical points is crucial to understand. In the context of limits, these functions sometimes lead to indeterminate forms or require special identities and techniques for evaluation.
For example, with the limit \( \lim_{x \rightarrow 0} \frac{\cos x - 1}{3x} \), understanding the behavior of \( \cos x \) near \( x = 0 \) is key. We know that \( \cos 0 = 1 \), so \( \cos x - 1 \) approaches zero as \( x \) approaches zero. This setup makes it eligible for applying L'Hôpital's Rule.
Other crucial trigonometric limits include those involving \( \sin x \) and \( \tan x \), such as \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \). Analyzing these limits helps in simplifying more complex problems.