The first problem in your homework is:

$\lim_{x \to 0} \frac{2x \cot^2(6x)}{\csc(12x)}$

Solution:

1. Rewrite using trigonometric identities:
   • Recall that $\csc(y) = \frac{1}{\sin(y)}$, so we can rewrite $\csc(12x)$ as $\frac{1}{\sin(12x)}$.
   • Also, $\cot(y) = \frac{\cos(y)}{\sin(y)}$, so $\cot^2(6x) = \frac{\cos^2(6x)}{\sin^2(6x)}$.

Thus, the limit becomes: $\lim_{x \to 0} \frac{2x \left(\frac{\cos^2(6x)}{\sin^2(6x)}\right)}{\frac{1}{\sin(12x)}} = \lim_{x \to 0} 2x \cdot \frac{\cos^2(6x) \sin(12x)}{\sin^2(6x)}$

2. Simplifying using small-angle approximations: For $x$ near 0, the following approximations hold:

   • $\sin(kx) \approx kx$,
   • $\cos(kx) \approx 1$.

   Using these:

   • $\sin(12x) \approx 12x$,
   • $\sin(6x) \approx 6x$,
   • $\cos(6x) \approx 1$.

Substituting these into the expression: $\lim_{x \to 0} 2x \cdot \frac{1 \cdot 12x}{(6x)^2} = \lim_{x \to 0} 2x \cdot \frac{12x}{36x^2}$

3. Simplify the limit expression: Cancel out the $x$ terms: $\lim_{x \to 0} 2x \cdot \frac{12}{36x} = \lim_{x \to 0} \frac{24}{36} = \frac{2}{3}$

Final Answer:

$\boxed{\frac{2}{3}}$

Would you like more details or further explanations?

5 Related Questions:

1. How do small-angle approximations help in solving limits?
2. What is the difference between $\sin(x)$ and $\cos(x)$ for small values of $x$?
3. Why is $\csc(x)$ important in trigonometric limits?
4. Can you explain more about L'Hopital's Rule for limits involving indeterminate forms?
5. How does the limit change if the angle inside sine or cosine is modified?

Tip:

Always remember small-angle approximations ($\sin(x) \approx x$, $\cos(x) \approx 1$) when dealing with trigonometric limits at $x = 0$!