1. **State the problem:** We want to estimate the limit of the function $$f(x) = \frac{2x}{x - 1}$$ as $$x$$ approaches infinity, i.e., $$\lim_{x \to \infty} \frac{2x}{x - 1}$$. 2. **Recall the rule for limits at infinity for rational functions:** When $$x$$ approaches infinity, the behavior of a rational function $$\frac{P(x)}{Q(x)}$$ depends on the degrees of the polynomials $$P(x)$$ and $$Q(x)$$. - If degrees are equal, the limit is the ratio of the leading coefficients. - If degree of numerator is greater, the limit is infinity or negative infinity. - If degree of denominator is greater, the limit is zero. 3. **Identify degrees and leading coefficients:** - Numerator: $$2x$$ has degree 1, leading coefficient 2. - Denominator: $$x - 1$$ has degree 1, leading coefficient 1. 4. **Apply the rule:** Since degrees are equal (both 1), the limit is the ratio of leading coefficients: $$\lim_{x \to \infty} \frac{2x}{x - 1} = \frac{2}{1} = 2$$ 5. **Confirm by dividing numerator and denominator by $$x$$:** $$\lim_{x \to \infty} \frac{2x}{x - 1} = \lim_{x \to \infty} \frac{\cancel{x} \cdot 2}{\cancel{x} \cdot 1 - \frac{1}{x}} = \lim_{x \to \infty} \frac{2}{1 - \frac{1}{x}}$$ As $$x \to \infty$$, $$\frac{1}{x} \to 0$$, so: $$\lim_{x \to \infty} \frac{2}{1 - 0} = 2$$ **Final answer:** $$\boxed{2}$$