Problem 52 - A Graphical Approach to Precalculus with Limits

Question

Determine each limit, if it exists. $$\lim _{x \rightarrow 0} \frac{\cos x+2 \sin x-1}{3 x}$$

Step-by-Step Solution

Answer

The limit is \( \frac{2}{3} \).

Step 1: Identify the Limit Form

We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\cos x+2 \sin x-1}{3 x} \). First, let's identify if it's an indeterminate form. By substituting \( x = 0 \), we get \( \frac{\cos 0 + 2\sin 0 - 1}{3 \times 0} = \frac{1 - 1}{0} = \frac{0}{0} \), which is an indeterminate form.

Step 2: Use L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we apply L'Hôpital's Rule, which states we can differentiate the numerator and the denominator separately: \[ \lim _{x \rightarrow 0} \frac{\cos x+2 \sin x-1}{3 x} = \lim _{x \rightarrow 0} \frac{-\sin x + 2\cos x}{3}. \]

Step 3: Evaluate the Limit After Differentiation

Now, evaluate the limit of the derivative: \[ \lim _{x \rightarrow 0} \frac{-\sin x + 2\cos x}{3}. \] Substitute \( x = 0 \), we have: \[ \frac{-\sin 0 + 2\cos 0}{3} = \frac{0 + 2 \cdot 1}{3} = \frac{2}{3}. \]

Step 4: Conclusion: Limit Value

Thus, the limit value is \( \frac{2}{3} \).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms occur when direct substitution in a limit results in an expression that does not immediately reveal useful information. For example, when substituting into a function, you might get forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( \infty - \infty \). These are called indeterminate because they don't clearly indicate a finite value or specific behavior of the function as you approach a certain point.
Detecting an indeterminate form is vital, as it suggests that further analysis is needed to understand the behavior of a function. In the provided exercise, substituting \(x = 0\) into the function \( \frac{\cos x + 2\sin x - 1}{3x} \) gives \( \frac{0}{0} \). This means the result is undefined at first glance, but it opens the door to apply more advanced techniques like L'Hôpital's Rule to resolve the limit.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for dealing with limits that present indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter these forms, the rule suggests that you can take the derivative of the numerator and the derivative of the denominator independently, then evaluate the limit of their quotient.

• The rule is applied only if both the original numerator and denominator approach zero or both approach infinity as the limit approaches a particular value.
• It is crucial to ensure that after applying L'Hôpital's Rule, the new limit is no longer an indeterminate form or can be resolved with another application of the rule.

In our example case, the function \( \frac{\cos x + 2\sin x - 1}{3x} \) transformed into \( \frac{-\sin x + 2\cos x}{3} \) after applying L'Hôpital's Rule. Evaluating this new function at \(x = 0\) then yielded a clear result of \( \frac{2}{3} \). This process shows how L'Hôpital's Rule helps us handle initial uncertainties and find meaningful values of limits.

Trigonometric Limits

Trigonometric limits can often seem daunting because of the periodic nature and complexity of trigonometric functions such as sine and cosine. However, they follow the same rules of calculus and often appear in problems involving indeterminate forms.

• Key limits to remember include \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \), which are frequently used in limit computations involving trigonometric functions.
• Understanding the derivatives of basic trigonometric functions can assist in using L'Hôpital's Rule effectively. For example, knowing that \( \frac{d}{dx} (\sin x) = \cos x \) and \( \frac{d}{dx} (\cos x) = -\sin x \) is essential.

In the provided exercise, evaluating \( \lim _{x \rightarrow 0} \frac{-\sin x + 2\cos x}{3} \) required recognizing these basic derivative rules to reach the conclusion that the original limit is \( \frac{2}{3} \). The reliance on derivatives showcases the interplay between trigonometric functions and calculus in resolving limits.