Problem 16
Question
Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{1-\cos ^{2} x}{x} $$
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Recognize the Trigonometric Identity
The given limit \( \lim _{x \rightarrow 0} \frac{1-\cos ^{2} x}{x} \) can be simplified using the Pythagorean identity for cosine: \( \sin^2 x = 1 - \cos^2 x \). Therefore, we can rewrite the expression as \( \frac{\sin^2 x}{x} \).
2Step 2: Apply Standard Limit
Recall the standard limit \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \). Although we have \( \sin^2 x \), we need the expression to match the standard limit format. Consider rewriting \( \sin^2 x \) as \( (\sin x) \cdot (\sin x) \).
3Step 3: Factor and Simplify the Limit Expression
Split \( \frac{\sin^2 x}{x} \) into two separate fractions: \( \sin x \cdot \frac{\sin x}{x} \). This allows us to apply the standard limit separately.
4Step 4: Evaluate the Split Limit
Use the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Thus, the limit \( \lim_{x \to 0} [\sin x \cdot \frac{\sin x}{x}] = \lim_{x \to 0} \sin x \cdot \lim_{x \to 0} \frac{\sin x}{x} = 0 \cdot 1 = 0 \).
5Step 5: Conclusion
Thus, the original limit \( \lim_{x \to 0} \frac{1-\cos^2 x}{x} = 0 \).
Key Concepts
Pythagorean Identity in TrigonometryStandard Limit Involving SineLimit Evaluation Technique
Pythagorean Identity in Trigonometry
The Pythagorean identity is a fundamental concept in trigonometry. It relates the sine and cosine functions, giving us a way to express one in terms of the other. This identity states:
- \( \sin^2 x + \cos^2 x = 1 \)
Standard Limit Involving Sine
In calculus, there is a well-known standard limit involving the sine function that is extremely handy when dealing with problems that involve trigonometric functions approaching zero. This standard limit is given by:
- \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
Limit Evaluation Technique
Evaluating limits, especially trigonometric ones, often involves a series of clever manipulations to bring an expression into a more familiar form. For example, in the given exercise, after using the Pythagorean identity, we needed to split the expression \( \frac{\sin^2 x}{x} \) into two parts:
- \( \sin x \) and \( \frac{\sin x}{x} \)
- \( \lim_{x \to 0} \sin x = 0 \)
- \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
Other exercises in this chapter
Problem 15
Evaluate the limits. $$ \lim _{x \rightarrow \infty} \frac{e^{x}}{e^{x}+2} $$
View solution Problem 15
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow \pi / 2} \sin (2 x) $$
View solution Problem 16
Use the formal definition of limits to prove each statement. $$ \lim _{x \rightarrow 3} \frac{1}{(x-3)^{2}}=\infty. $$
View solution Problem 16
Use the Intermediate Value Theorem to show that \(e^{x}-\) \(e^{-1.5 x}-1=0\) has a solution in the interval \((0,1) .\) Use a spreadsheet to calculate the valu
View solution