Answer the following questions.

(1) For $x\geq 0$, show that $x-\frac{x^3}{6}\leq \sin x\leq x.$

(2) For $x\geq 0$, show that $\frac{x^3}{3}-\frac{x^5}{30}\leq \int_0^x t\sin t\ dt\leq \frac{x^3}{3}.$

(3) Find the limit 

\[\lim_{x\rightarrow 0} \frac{\sin x-x\cos x}{x^3}.\]
1. **Proving \( x - \frac{x^3}{6} \leq \sin x \leq x \) for \( x \geq 0 \):**

   We start with the known inequality for the cosine function:
   \[
   \cos x \leq 1 \quad \text{for all } x.
   \]
   Integrating both sides from 0 to \( x \):
   \[
   \int_0^x \cos t \, dt \leq \int_0^x 1 \, dt,
   \]
   which gives:
   \[
   \sin x \leq x.
   \]

   Next, we use the Taylor series expansion for \(\cos x\):
   \[
   \cos x = 1 - \frac{x^2}{2} + O(x^4).
   \]
   Therefore,
   \[
   1 - \frac{x^2}{2} \leq \cos x \leq 1.
   \]
   Integrating from 0 to \( x \):
   \[
   \int_0^x \left(1 - \frac{t^2}{2}\right) dt \leq \int_0^x \cos t \, dt \leq \int_0^x 1 \, dt,
   \]
   which simplifies to:
   \[
   x - \frac{x^3}{6} \leq \sin x \leq x.
   \]

2. **Proving \( \frac{x^3}{3} - \frac{x^5}{30} \leq \int_0^x t \sin t \, dt \leq \frac{x^3}{3} \) for \( x \geq 0 \):**

   From the first part, we have:
   \[
   x - \frac{x^3}{6} \leq \sin x \leq x.
   \]
   Multiplying by \( t \):
   \[
   t \left( x - \frac{x^3}{6} \right) \leq t \sin t \leq t x.
   \]
   Integrating from 0 to \( x \):
   \[
   \int_0^x t \left( x - \frac{x^3}{6} \right) dt \leq \int_0^x t \sin t \, dt \leq \int_0^x t x \, dt.
   \]
   Simplifying the integrals:
   \[
   \int_0^x t x \, dt = x \int_0^x t \, dt = x \left[ \frac{t^2}{2} \right]_0^x = \frac{x^3}{2},
   \]
   and
   \[
   \int_0^x t \left( x - \frac{x^3}{6} \right) dt = \left( x - \frac{x^3}{6} \right) \int_0^x t \, dt = \left( x - \frac{x^3}{6} \right) \frac{x^2}{2} = \frac{x^3}{2} - \frac{x^5}{12}.
   \]
   Therefore,
   \[
   \frac{x^3}{3} - \frac{x^5}{30} \leq \int_0^x t \sin t \, dt \leq \frac{x^3}{3}.
   \]

3. **Finding the limit \( \lim_{x \rightarrow 0} \frac{\sin x - x \cos x}{x^3} \):**

   Using the Taylor series expansions:
   \[
   \sin x = x - \frac{x^3}{6} + O(x^5),
   \]
   and
   \[
   x \cos x = x \left( 1 - \frac{x^2}{2} + O(x^4) \right) = x - \frac{x^3}{2} + O(x^5).
   \]
   Therefore,
   \[
   \sin x - x \cos x = \left( x - \frac{x^3}{6} + O(x^5) \right) - \left( x - \frac{x^3}{2} + O(x^5) \right) = -\frac{x^3}{6} + \frac{x^3}{2} + O(x^5) = \frac{x^3}{3} + O(x^5).
   \]
   Dividing by \( x^3 \):
   \[
   \frac{\sin x - x \cos x}{x^3} = \frac{\frac{x^3}{3} + O(x^5)}{x^3} = \frac{1}{3} + O(x^2).
   \]
   Taking the limit as \( x \to 0 \):
   \[
   \lim_{x \rightarrow 0} \frac{\sin x - x \cos x}{x^3} = \frac{1}{3}.
   \]

The final answer is \( \boxed{ \frac{1}{3} } \).