Using $\epsilon$ and $\delta$ to prove that $\lim_{x \to 0}\frac{\tan x}{x}=1$

Question

I tried $$ \left | \frac{\tan x}{x}-1 \right | <1+\left | \frac{\sin x}{x} \right |\left | \frac{1}{\cos x} \right |<1+\frac{\delta^2}{2}\left | \frac{1}{\cos x} \right | $$

and I failed to deal with $\left | \frac{1}{\cos x} \right |$.

Is there an easier way, or something with this way ?

Answer 1

Use $\tan x=\frac{\sin x}{\cos x}$ and $\lim_{x\to 0}\frac{\sin x}{x}=1$.