Slope of the tangent line via the limit definition

First, $f(3) = \dfrac{1}{3-1} = \dfrac{1}{2}$. The difference quotient is $\frac{f(x)-f(3)}{x-3} = \frac{\frac{1}{x-1} - \frac{1}{2}}{x-3}.$ Combine the numerator over a common denominator: $\frac{1}{x-1} - \frac{1}{2} = \frac{2 - (x-1)}{2(x-1)} = \frac{3-x}{2(x-1)}$.

So the difference quotient becomes $\frac{3-x}{2(x-1)(x-3)} = \frac{-(x-3)}{2(x-1)(x-3)} = \frac{-1}{2(x-1)},$ after canceling the factor $(x-3)$, valid for $x \neq 3$. Taking the limit as $x \to 3$ gives $f'(3) = \dfrac{-1}{2(3-1)} = -\dfrac{1}{4}$.

Since $f(1)$ is not defined (division by zero), the point $(1, f(1))$ does not exist on the graph, so there is no function value at $x=1$ to which secant lines can converge. As $x \to 1$, $f(x) \to \pm\infty$, producing a vertical asymptote rather than a well-defined point.

Because $f$ is not even defined at $x=1$, it cannot be continuous there, and differentiability requires the function to be defined (and continuous) at that point. Therefore $f$ is not differentiable at $x=1$; the domain restriction $x \neq 1$ directly causes this failure.