$f(x) = \dfrac{(x-1)(x-3)}{(x-2)^2}$

Consider the function above, what is the value of $\displaystyle \lim_{x \to \infty} f(x)$ ?

The answer is 1.

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$f(x) = \dfrac{(x-1)(x-3)}{(x-2)^2} = \dfrac{x^2 - 4x +3}{x^2 - 4x +4} = 1 - \dfrac{1}{x^2 - 4x +4}$ .

As expanded above, when $x$ approaches infinity, the value of $\dfrac{1}{x^2 - 4x +4}$ will reach $0$ , and as shown in the graph below, the limit of this function is, therefore, $1$ .