Math Problem Statement
Find the limit lim (x→0) [(csc x - cos x cot x) / 2x]
Solution
The limit to be evaluated is:
We will solve this by simplifying the expression and using known limit properties.
- Recall trigonometric identities:
Using these, the expression becomes:
= \lim_{x \to 0} \frac{\frac{1 - \cos^2 x}{\sin x}}{2x}$$ Now, recall that $$1 - \cos^2 x = \sin^2 x$$, so the expression simplifies to: $$\lim_{x \to 0} \frac{\sin x}{2x}$$ 2. **Evaluate the limit:** $$\lim_{x \to 0} \frac{\sin x}{2x}$$ We know from standard limits that $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. Therefore, we have: $$\lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2}$$ Thus, the value of the limit is: $$\boxed{\frac{1}{2}}$$ Would you like further clarification, or do you have any additional questions? Here are 5 related questions: 1. How does the identity $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ arise? 2. What are other common trigonometric limits that are useful in calculus? 3. Can you explain how L'Hopital's Rule applies to limits like this? 4. How do you derive the Taylor expansion for $$\sin x$$ and $$\cos x$$? 5. What happens to the limit if we change $$2x$$ in the denominator to another function? **Tip:** Memorizing standard trigonometric limits such as $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ and $$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$ can significantly simplify solving these types of problems.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Trigonometry
Limits
Indeterminate Forms
Formulas
csc x = 1 / sin x
cot x = cos x / sin x
lim (x→0) [sin x / x] = 1
Theorems
Trigonometric Limit Theorems
Limit Simplification
Suitable Grade Level
College Calculus or AP Calculus