Consider a curve $C$ on the $x$-$y$ plane expressed by $x=\tan \theta ,\ y=\frac{1}{\cos \theta}\left (0\leq \theta <\frac{\pi}{2}\right)$.
 For a constant $t>0$, let the line $l$ pass through the point $P(t,\ 0)$ and is perpendicular to the $x$-axis,intersects with the curve $C$ at $Q$. Denote by $S_1$ the area of the figure bounded by the curve $C$, the $x$-axis, the $y$-axis and the line $l$, and denote by $S_2$ the area of $\triangle{OPQ}$. Find $\lim_{t\to\infty} \frac{S_1-S_2}{\ln t}.$
1. **Express the curve \( C \) in terms of \(\theta\):**
   \[
   x = \tan \theta, \quad y = \frac{1}{\cos \theta} \quad \left(0 \leq \theta < \frac{\pi}{2}\right)
   \]
   This implies the relationship:
   \[
   x^2 - y^2 = \tan^2 \theta - \frac{1}{\cos^2 \theta} = \tan^2 \theta - \sec^2 \theta = \tan^2 \theta - (\tan^2 \theta + 1) = -1
   \]
   Therefore, the curve \( C \) can be expressed as:
   \[
   x^2 - y^2 = -1
   \]

2. **Determine the intersection point \( Q \) of the line \( l \) and the curve \( C \):**
   The line \( l \) passing through \( P(t, 0) \) and perpendicular to the \( x \)-axis is given by:
   \[
   x = t
   \]
   Substituting \( x = t \) into the curve equation \( x^2 - y^2 = -1 \):
   \[
   t^2 - y^2 = -1 \implies y^2 = t^2 + 1 \implies y = \sqrt{t^2 + 1}
   \]
   Thus, the intersection point \( Q \) is:
   \[
   Q(t, \sqrt{t^2 + 1})
   \]

3. **Calculate the area \( S_1 \):**
   \( S_1 \) is the area bounded by the curve \( C \), the \( x \)-axis, the \( y \)-axis, and the line \( l \). This area can be found by integrating the function \( y = \sqrt{x^2 + 1} \) from \( x = 0 \) to \( x = t \):
   \[
   S_1 = \int_{0}^{t} \sqrt{x^2 + 1} \, dx
   \]

4. **Calculate the area \( S_2 \):**
   \( S_2 \) is the area of the triangle \( \triangle OPQ \). The vertices of the triangle are \( O(0, 0) \), \( P(t, 0) \), and \( Q(t, \sqrt{t^2 + 1}) \). The area of the triangle is given by:
   \[
   S_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times \sqrt{t^2 + 1}
   \]

5. **Evaluate the integral for \( S_1 \):**
   To find \( S_1 \), we use the integral:
   \[
   \int_{0}^{t} \sqrt{x^2 + 1} \, dx
   \]
   Using the substitution \( x = \sinh u \), \( dx = \cosh u \, du \):
   \[
   \int_{0}^{t} \sqrt{x^2 + 1} \, dx = \int_{0}^{\sinh^{-1}(t)} \cosh^2 u \, du = \int_{0}^{\sinh^{-1}(t)} \frac{1 + \cosh(2u)}{2} \, du
   \]
   \[
   = \frac{1}{2} \int_{0}^{\sinh^{-1}(t)} 1 \, du + \frac{1}{2} \int_{0}^{\sinh^{-1}(t)} \cosh(2u) \, du
   \]
   \[
   = \frac{1}{2} \sinh^{-1}(t) + \frac{1}{4} \sinh(2u) \bigg|_{0}^{\sinh^{-1}(t)}
   \]
   \[
   = \frac{1}{2} \sinh^{-1}(t) + \frac{1}{4} \sinh(2 \sinh^{-1}(t))
   \]
   Using the identity \( \sinh(2u) = 2 \sinh(u) \cosh(u) \):
   \[
   = \frac{1}{2} \sinh^{-1}(t) + \frac{1}{4} \cdot 2 \sinh(\sinh^{-1}(t)) \cosh(\sinh^{-1}(t))
   \]
   \[
   = \frac{1}{2} \sinh^{-1}(t) + \frac{1}{2} t \sqrt{t^2 + 1}
   \]

6. **Find the limit:**
   \[
   \lim_{t \to \infty} \frac{S_1 - S_2}{\ln t}
   \]
   \[
   S_1 = \frac{1}{2} \sinh^{-1}(t) + \frac{1}{2} t \sqrt{t^2 + 1}
   \]
   \[
   S_2 = \frac{1}{2} t \sqrt{t^2 + 1}
   \]
   \[
   S_1 - S_2 = \frac{1}{2} \sinh^{-1}(t)
   \]
   \[
   \lim_{t \to \infty} \frac{\frac{1}{2} \sinh^{-1}(t)}{\ln t}
   \]
   Using the asymptotic behavior \( \sinh^{-1}(t) \sim \ln(2t) \) as \( t \to \infty \):
   \[
   \lim_{t \to \infty} \frac{\frac{1}{2} \ln(2t)}{\ln t} = \lim_{t \to \infty} \frac{\frac{1}{2} (\ln 2 + \ln t)}{\ln t} = \lim_{t \to \infty} \frac{\frac{1}{2} \ln 2 + \frac{1}{2} \ln t}{\ln t} = \frac{1}{2}
   \]

The final answer is \(\boxed{\frac{1}{2}}\).