Evaluate the limit: lim (x->0) [sin^4(2x) * cot^2(x)] / (8x^2)
Understand the Problem
The question asks to evaluate a limit of a trigonometric function as x approaches 0. The provided solution attempts to simplify the expression and use standard limit results to find the final answer. We need to verify the correctness of the steps.
Answer
2
Answer for screen readers
2
Steps to Solve
-
Rewrite $\cot^2(x)$ Rewrite $\cot^2(x)$ as $\frac{\cos^2(x)}{\sin^2(x)}$. This gives us: $$ \lim_{x \to 0} \frac{\sin^4(2x) \cot^2(x)}{8x^2} = \lim_{x \to 0} \frac{\sin^4(2x) \cos^2(x)}{8x^2 \sin^2(x)} $$
-
Multiply by $\frac{(2x)^4}{(2x)^4}$ and $\frac{x^2}{x^2}$ Multiply the expression by $\frac{(2x)^4}{(2x)^4} \cdot \frac{x^2}{x^2}$. $$ \lim_{x \to 0} \frac{\sin^4(2x) \cos^2(x)}{8x^2 \sin^2(x)} \cdot \frac{(2x)^4}{(2x)^4} \cdot \frac{x^2}{x^2} $$
-
Rearrange Terms Rearrange the terms to group $\frac{\sin^4(2x)}{(2x)^4}$ and $\frac{x^2}{\sin^2(x)}$ together. This yields: $$ \lim_{x \to 0} \frac{\sin^4(2x)}{(2x)^4} \cdot \frac{x^2}{\sin^2(x)} \cdot \frac{(2x)^4 \cos^2(x)}{8x^2 x^2} $$
-
Simplify the last term Simplify $\frac{(2x)^4 \cos^2(x)}{8x^2 x^2}$. $$ \frac{(2x)^4 \cos^2(x)}{8x^2 x^2} = \frac{16x^4 \cos^2(x)}{8x^4} = 2\cos^2(x) $$
-
Evaluate the limit Evaluate the limit as $x \to 0$. We know that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, so $\lim_{x \to 0} \frac{\sin^4(2x)}{(2x)^4} = 1$ and $\lim_{x \to 0} \frac{x^2}{\sin^2(x)} = 1$. Also, $\lim_{x \to 0} \cos^2(x) = 1$. Therefore, $$ \lim_{x \to 0} \frac{\sin^4(2x)}{(2x)^4} \cdot \frac{x^2}{\sin^2(x)} \cdot 2\cos^2(x) = 1 \cdot 1 \cdot 2 \cdot 1 = 2 $$
2
More Information
The limit evaluates to 2.
Tips
The error in the original solution is in the rearrangement of terms in step 3. Specifically, the term $\frac{\cos^2(x)}{8x^2}$ should be $2\cos^2(x)$ after simplification. This changes the final answer.
AI-generated content may contain errors. Please verify critical information
