Evaluate $\lim_{x\rightarrow 0} \frac{\sin x}{x + \tan x} $ without L'Hopital I need help finding the the following limit: 
$$\lim_{x\rightarrow 0} \frac{\sin x}{x + \tan x} $$
I tried to simplify to:
$$ \lim_{x\rightarrow 0} \frac{\sin x \cos x}{x\cos x+\sin x} $$
but I don't know where to go from there. I think, at some point, you have to use the fact that $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$. Any help would be appreciated.
Thanks!
 A: $$
\frac{\sin x}{x + \tan x} = \frac{1}{\frac{x}{\sin x}+\frac{\tan x}{\sin x}} \to 1/2
$$
A: Turn it upside down. ${}{}{}{}{}{}$
A: Hint
Divide both numerator and denominator by $x$ and use the fact
$$\lim_{x\to0}\frac{\sin x}{x}=1$$
A: $$\lim_{x\rightarrow 0} \frac{\sin x}{x + \tan x} =\lim_{x\rightarrow 0} \frac{1}{\frac{x}{\sin x} + \frac{1}{cos x}} =1/2$$
A: $$\lim_{x\rightarrow 0}\frac{\sin x}{x+\tan x}=\lim_{x\rightarrow 0}\frac{\frac{\sin x}{x}}{1+\frac{\tan x}{x}}=\frac {1}{1+1}=\frac{1}{2}$$
A: Given that
$$
\lim_{x \to 0} \frac{\sin x}{x} = 1,
$$
we can calculate that
$$
\begin{align}
\lim_{x \to 0} \frac{\tan x}{x} &= \lim_{x \to 0} \frac{\sin x}{x \cos x} \\
&= \lim_{x \to 0} \frac{\sin x} {x} \cdot \lim_{x \to 0} \frac{1}{\cos x} \\
&= 1.
\end{align}
$$

Now, rewrite your limit and compute:
$$
\lim_{x \to 0} \frac{\sin x}{x + \tan x} = \lim_{x \to 0} \frac{\frac{\sin x}{x}}{1 + \frac{\tan x}{x}} = \frac{1}{2}.
$$
A: $$\sin x \sim x$$
$$\tan x \sim x$$
Hence
$$\frac{\sin x}{x + \tan x} \sim \frac{x}{x+x} = \frac{1}{2}$$
A: Look at the ratio of the Taylor series:
$\frac{\sin x}{x+\tan x} = \frac{x-\frac{x^3}{3!}+\cdots}{x + (x+\frac{x^3}{3}+\cdots)}$. Combine the $x$ terms in the denominator, divide out a factor of $x$, and you have
$\frac{1 - \frac{x^2}{2!}+\cdots}{2 + \frac{x^2}{3} + \cdots}$. Then you can let $x\rightarrow 0$ to get $\frac12$ because only the constant terms remain in the limit.
