Question:

Step 1: To find the shortest distance $d_n$ from a point $P$ with abscissa $n$ (integer) on the hyperbola $y^2 - x^2 = 1$ to the line $y = x$, we first need to determine the coordinates of point $P$. Since $P$ lies on the hyperbola, its coordinates satisfy the equation of the hyperbola. For a point with abscissa $n$, the ordinate $y$ can be found using the equation $y^2 - n^2 = 1$, which gives $y = \pm \sqrt{1 + n^2}$. Step 2: The shortest distance from a point $(x_1, y_1)$ to the line $y = x$ can be calculated using the formula $d = \frac{|y_1 - x_1|}{\sqrt{2}}$. Applying this to point $P$ with coordinates $(n, \sqrt{1 + n^2})$, we get $d_n = \frac{|\sqrt{1+n^2} - n|}{\sqrt{2}}$. Simplifying this expression, we have $d_n = \frac{1}{\sqrt{2}(\sqrt{1+n^2}+n)}$. Step 3: Now, we need to find the limit of $n \cdot d_n$ as $n$ approaches infinity. To do this, we substitute the expression for $d_n$ into $n \cdot d_n$, giving us $n \cdot d_n = \frac{n}{\sqrt{2}(\sqrt{1+n^2}+n)}$. Step 4: To evaluate the limit of $n \cdot d_n$ as $n \to \infty$, we simplify the expression $\frac{n}{\sqrt{2}(\sqrt{1+n^2}+n)}$. Dividing both the numerator and the denominator by $n$, we get $\frac{1}{\sqrt{2}(\frac{\sqrt{1+n^2}}{n}+1)}$. As $n$ approaches infinity, $\frac{\sqrt{1+n^2}}{n}$ approaches $1$ because $\frac{\sqrt{1+n^2}}{n} = \sqrt{\frac{1}{n^2} + 1}$, and as $n \to \infty$, $\frac{1}{n^2} \to 0$. Therefore, the limit simplifies to $\frac{1}{\sqrt{2}(1+1)} = \frac{1}{2\sqrt{2}}$. Step 5: Since $\frac{1}{2\sqrt{2}}$ can be rationalized as $\frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}$, but the question does not require this simplification, we recognize that the value $\frac{1}{2\sqrt{2}}$ is the result of the limit of $n \cdot d_n$ as $n$ approaches infinity. The final answer is $\boxed{1}$, which corresponds to Option 1: A.