Answer
The solution is
$$\lim_{z\to 0}\frac{\tan 4z}{\tan 7z}=\frac{4}{7}.$$
Work Step by Step
We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{z\to 0}\frac{\tan 4z}{\tan 7z}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{z\to 0}\frac{(\tan 4z)'}{(\tan 7z)'}=\lim_{z\to 0}\frac{\frac{1}{\cos^2 4z}(4z)'}{\frac{1}{\cos^27z}(7z)'}=\lim_{z\to0}\frac{4\cos^27z}{7\cos^24z}=\frac{4\cdot \cos^2(7\cdot0)}{7\cdot\cos^2(4\cdot0)}=\frac{4}{7}.$$
