Calculate $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x + \tan(x)}$ with out derivative?

Question

How do I calculate $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x + \tan(x)}$ without derivative?

I cant see the trigonometrical operations that will simply this limit...

Answer 1

Divide by $\sin x$ to get $$\frac{1}{\frac{x}{\sin x} + \sec x}$$

Since the limit of $\frac{x}{\sin x}$ is $1$, $\sec(0)$ is $1$, so the limit $= \frac{1}{2}$.

Answer 2

Use \begin{align*} \sin x &= x + o(x) \\ \tan x &= x + o(x) \end{align*} to get $$ \frac{\sin x}{x+\tan x} = \frac{x+o(x)}{2x+o(x)} = \frac{1+o(1)}{2+o(1)} \xrightarrow[x\to 0]{}\frac{1}{2}. $$

Answer 3

We can do the following: $$\lim_{x\to 0}\frac{\sin x}{x+\tan x}=\lim_{x\to 0}\frac{\frac{\sin x}{x}}{1+\frac{\tan x}{x}}=\lim_{x\to 0}\frac{\frac{\sin x}{x}}{1+\frac{\sin x}{x}\cdot{\frac{1}{\cos x}}}=\frac{1}{1+1\cdot 1}=\frac{1}{2}$$