Limit Tan Sin 0A125C

1. **State the problem:** We need to find the limit $$\lim_{t \to 0} \frac{\tan(6t)}{\sin(3t)}$$. 2. **Recall important limits and formulas:** - As $x \to 0$, $\tan x \approx x$ and $\sin x \approx x$. - This means $\lim_{x \to 0} \frac{\tan x}{x} = 1$ and $\lim_{x \to 0} \frac{\sin x}{x} = 1$. 3. **Rewrite the expression using these approximations:** $$\frac{\tan(6t)}{\sin(3t)} = \frac{\tan(6t)}{6t} \cdot \frac{6t}{3t} \cdot \frac{3t}{\sin(3t)}$$ 4. **Simplify the middle fraction:** $$\frac{6t}{3t} = \cancel{\frac{6\cancel{t}}{3\cancel{t}}} = 2$$ 5. **Apply the limits to the other two fractions:** - $$\lim_{t \to 0} \frac{\tan(6t)}{6t} = 1$$ - $$\lim_{t \to 0} \frac{3t}{\sin(3t)} = 1$$ 6. **Multiply all parts together:** $$1 \times 2 \times 1 = 2$$ 7. **Final answer:** $$\lim_{t \to 0} \frac{\tan(6t)}{\sin(3t)} = 2$$