evaluate the limit\n\\( \\lim _{x \\rightarrow 0} \\frac{\\tan x}{2 x} \\)

Explanation:

Step1: Use the identity $\tan x=\frac{\sin x}{\cos x}$

$$\lim_{x\rightarrow0}\frac{\tan x}{2x}=\lim_{x\rightarrow0}\frac{\sin x}{2x\cos x}$$

Step2: Split the limit

$$\lim_{x\rightarrow0}\frac{\sin x}{2x\cos x}=\frac{1}{2}\lim_{x\rightarrow0}\frac{\sin x}{x}\cdot\lim_{x\rightarrow0}\frac{1}{\cos x}$$

Step3: Use the well - known limit $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$

Since $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$ and $\lim_{x\rightarrow0}\cos x=\cos(0) = 1$ (by the continuity of the cosine function, $\lim_{x\rightarrow a}\cos x=\cos a$) $$\frac{1}{2}\lim_{x\rightarrow0}\frac{\sin x}{x}\cdot\lim_{x\rightarrow0}\frac{1}{\cos x}=\frac{1}{2}\times1\times\frac{1}{1}$$

Answer:

$\frac{1}{2}$