Evaluate $\lim_{x\to 0}\frac{x-\sin x}{x\sin x}$ without to use L'Hopital 
Evaluate $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}$$ Without L'Hopital's Rule 

$$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{x(1-\frac{\sin x}{x})}{x\sin x}=\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}$$
But I can not find a way to deal with $\sin x$ that does not result with a limit of the type $"\frac{0}{0}"$
 A: A rigorous treatment of trig functions would necessitate the use of power series, when this becomes trivial (with all the tools that come along with the theory of converging power series). Let me proffer the following argument relying on looking up things from a picture. 
The function is odd, so it suffices to handle the case $x>0$. Let's look at the part of the unit circle: $O=(0,0)$, $A=(1,0)$, $P=(\cos x, \sin x)$. 

The triangle $\Delta OAP$ has area $\dfrac12\sin x$.
That triangle is contained in the sector $\angle OAP$ with area $\dfrac12 x$
($x$ is the length of the arc $AP$). Therefore $0<\sin x<x$. Presumably all are familiar with this estimate. I will use it below without mentioning it explicitly. 
Furthermore, the thin circular segment in between the arc $AP$ and the line segment $AP$ has area $(x-\sin x)/2$.
The line segment $AP$ has length $2\sin(x/2)$, and the maximum height of the circular segment is obviously (along the red line in the image) $h=1-\cos(x/2)$. Therefore the area of the segment is bounded from above by $2h\sin(x/2)$. This implies the inequalities
$$
0<\frac{x-\sin x}{x\sin x}<\frac{4\sin (x/2)(1-\cos(x/2))}{x\sin x}.
$$
Let's work on that upper bound. In the denominator we can write the sine as
$$
\sin x=2\sin(x/2)\cos(x/2).
$$
Multiplying both the numerator and the denominator by $1+\cos(x/2)$ then gives the upper bound
$$
\begin{aligned}
\frac{4\sin (x/2)(1-\cos(x/2))}{x\sin x}&=\frac{4\sin^3(x/2)}{2x\sin(x/2)\cos(x/2)(1+\cos(x/2))}\\
&<\frac{x}{2\cos(x/2)(1+\cos(x/2))}\\
&<x
\end{aligned}
$$
for all $x\in(0,\pi/3)$ (when $2\cos(x/2)$ exceeds $1$).
The squeeze theorem then implies that the limit as $x\to0+$ is equal to zero. 
A: Let $0<x<\pi/2.$ Consider the sector within the unit circle with vertices $(0,0),(1,0),(\cos x,\sin x);$ its area is $x/2.$ That area is smaller than the area of the right triangle with vertices $(0,0),(1,0),(1,\tan x),$ whose area is $(\tan x)/2.$ (Good to draw a picture.) This gives the inequality $x < \tan x.$ Thus
$$0< x-\sin x < \tan x - \sin x = \tan x (1-\cos x)$$ $$ < \tan x (1-\cos^2 x) = \tan x \cdot \sin^2 x.$$
Therefore
$$0< \frac{x-\sin x}{x^2} < \tan x \frac{\sin^2 x}{x^2}.$$
As $x\to 0^+,$ $\tan x \to 0$ and $(\sin^2 x)/x^2 \to 1.$ Thus our limit, from the right, is $0.$ Because our function is odd, the limit is also $0$ from the left, and we're done.
A: Using a taylor expansion for $\sin(x)$,  you can find the limit.
Hint: $\sin(x) = x − \frac{x^3}{3!} + \frac{x^5}{5!} −O(x^7)$
A: Given your steps, without derivates and Taylor,
$$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{x(1-\frac{\sin x}{x})}{x\sin x}=\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}$$
an knowing that
$$\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\frac16$$
you have:
$$\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}=\lim_{x\to 0}\frac{x^2}{x^2}\left(\frac{1-\frac{\sin x}{x}}{\sin x}\right)=\lim_{x\to 0}\left[\underbrace{\frac{1-\tfrac{\sin x}{x}}{x^2}}_{\to \frac 16}\left(\underbrace{\frac{x}{\sin x}}_{\to 1}\right)\right]\, x=0$$
