For positive constant $a$, let $C: y=\frac{a}{2}(e^{\frac{x}{a}}+e^{-\frac{x}{a}})$. Denote by $l(t)$ the length of the part $a\leq y\leq t$ for $C$ and denote by $S(t)$ the area of the part bounded by the line $y=t\ (a<t)$ and $C$. Find $\lim_{t\to\infty} \frac{S(t)}{l(t)\ln t}.$
1. Given the curve \( C: y = \frac{a}{2} \left( e^{\frac{x}{a}} + e^{-\frac{x}{a}} \right) \), we can rewrite it using the hyperbolic cosine function:
   \[
   y = a \cosh \left( \frac{x}{a} \right)
   \]
   where \(\cosh(z) = \frac{e^z + e^{-z}}{2}\).

2. Let \( b(t) \) be such that \( y(b(t)) = t \) with \( b(t) \ge 0 \). This means:
   \[
   t = a \cosh \left( \frac{b(t)}{a} \right)
   \]

3. The length \( l(t) \) of the part of the curve where \( a \leq y \leq t \) is given by:
   \[
   l(t) = \int_{-b(t)}^{b(t)} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
   \]
   Since \( y = a \cosh \left( \frac{x}{a} \right) \), we have:
   \[
   \frac{dy}{dx} = \sinh \left( \frac{x}{a} \right)
   \]
   Therefore:
   \[
   l(t) = \int_{-b(t)}^{b(t)} \sqrt{1 + \sinh^2 \left( \frac{x}{a} \right)} \, dx = \int_{-b(t)}^{b(t)} \cosh \left( \frac{x}{a} \right) \, dx
   \]
   Using the symmetry of the integrand:
   \[
   l(t) = 2 \int_0^{b(t)} \cosh \left( \frac{x}{a} \right) \, dx
   \]
   Integrating:
   \[
   l(t) = 2a \sinh \left( \frac{b(t)}{a} \right)
   \]
   Since \( \cosh \left( \frac{b(t)}{a} \right) = \frac{t}{a} \), we have:
   \[
   \sinh \left( \frac{b(t)}{a} \right) = \sqrt{\left( \frac{t}{a} \right)^2 - 1} = \frac{\sqrt{t^2 - a^2}}{a}
   \]
   Thus:
   \[
   l(t) = 2 \sqrt{t^2 - a^2}
   \]

4. The area \( S(t) \) of the part bounded by the line \( y = t \) and the curve \( C \) is given by:
   \[
   S(t) = \int_{-b(t)}^{b(t)} (t - y(x)) \, dx
   \]
   Using the symmetry of the integrand:
   \[
   S(t) = 2 \int_0^{b(t)} (t - a \cosh \left( \frac{x}{a} \right)) \, dx
   \]
   Integrating:
   \[
   S(t) = 2 \left[ tx - a^2 \sinh \left( \frac{x}{a} \right) \right]_0^{b(t)} = 2 \left( tb(t) - a^2 \sinh \left( \frac{b(t)}{a} \right) \right)
   \]
   Substituting \( \sinh \left( \frac{b(t)}{a} \right) \):
   \[
   S(t) = 2 \left( tb(t) - a \sqrt{t^2 - a^2} \right)
   \]

5. For large \( t \), we approximate \( b(t) \) using the fact that \( \cosh \left( \frac{b(t)}{a} \right) \approx \frac{1}{2} \left( e^{\frac{b(t)}{a}} + e^{-\frac{b(t)}{a}} \right) \approx \frac{1}{2} e^{\frac{b(t)}{a}} \) for large \( t \):
   \[
   \frac{t}{a} \approx \frac{1}{2} e^{\frac{b(t)}{a}}
   \]
   Solving for \( b(t) \):
   \[
   e^{\frac{b(t)}{a}} \approx \frac{2t}{a} \implies \frac{b(t)}{a} \approx \ln \left( \frac{2t}{a} \right) \implies b(t) \approx a \ln \left( \frac{2t}{a} \right)
   \]

6. Finally, we find the limit:
   \[
   \lim_{t \to \infty} \frac{S(t)}{l(t) \ln t} = \lim_{t \to \infty} \frac{2 \left( tb(t) - a \sqrt{t^2 - a^2} \right)}{2 \sqrt{t^2 - a^2} \ln t} = \lim_{t \to \infty} \frac{tb(t) - a \sqrt{t^2 - a^2}}{\sqrt{t^2 - a^2} \ln t}
   \]
   For large \( t \):
   \[
   \sqrt{t^2 - a^2} \approx t
   \]
   Thus:
   \[
   \lim_{t \to \infty} \frac{tb(t) - at}{t \ln t} = \lim_{t \to \infty} \frac{b(t) - a}{\ln t} = \lim_{t \to \infty} \frac{a \ln \left( \frac{2t}{a} \right) - a}{\ln t} = \lim_{t \to \infty} \frac{a \left( \ln 2 + \ln t - \ln a \right) - a}{\ln t} = a
   \]

\(\blacksquare\)

The final answer is \( \boxed{ a } \)