Evaluate $ \lim_{x\to 0} \frac{\tan(4x)}{\sin(7x)}$ Evaluate $$ \lim_{x\to 0} \  \frac{\tan(4x)}{\sin(7x)}$$
I am stuck after I convert tan(4x) into sin (4x) / cosine(4x)  
 A: Recall that $\lim_{x\to0} \sin(x)/x = 1$ and $\lim_{x\to 0} \tan(x)/x = 1$.
Then $$\lim_{x\to 0} \frac{\tan(4x)}{\sin(7x)} = \lim_{x\to 0} \frac{7}{7} \cdot \frac{4x}{4x} \cdot \frac{\tan(4x)}{\sin(7x)} = \lim_{x\to 0} \frac{4}{7} \frac{\tan(4x)}{4x} \cdot \frac{7x}{\sin(7x)} = \frac47 \cdot 1 \cdot 1 = \frac47.$$

Once we accept that $\lim_{x\to0} \frac{\sin(x)}{x} =1$ it immediately follows that $\lim_{x\to0} \frac{\tan(x)}{x} = 1$, since $$\lim_{x\to0} \frac{\tan(x)}{x} = \lim_{x\to0} \frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)} = 1 \cdot 1 = 1$$
A: Other answers are correct and valid. If you know l'Hôpital's rule, there's another way. Observe:
$$\lim_{x \to 0} \frac{\tan(4x)}{\sin(7x)} = \lim_{x \to 0} \frac{\frac{d}{dx}\tan(4x)}{\frac{d}{dx}\sin(7x)} = \lim_{x \to 0} \frac{4\sec^2(4x)}{7\cos(7x)} = \frac{4}{7}\frac{\sec^2(0)}{\cos(0)} = \frac{4}{7}\frac{1}{1} = \frac{4}{7}$$
A: Hints: $\lim_{x\to 0} \cos(4x) = 1$, and
\begin{equation*}
  \lim_{x\to 0} \frac{\sin(4x)}{\sin(7x)} = \frac{4}{7}\lim_{x\to 0}\left(\frac{\sin(4x)}{4x}\cdot\frac{7x}{\sin(7x)}\right).
\end{equation*}
