Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 21

PROBLEM:

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SOLUTION:

\begin{align*} \displaystyle \lim\limits_{x\to 2}\left(\displaystyle \frac{f\left(x\right)-f\left(2\right)}{x-2}\right) & =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-\left(2^2-2\cdot 2+3\right)}{x-2}\right) \\ \\ & =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-3}{x-2}\right)\\ \\ & =\lim\limits_{x\to 2}\left(\displaystyle \frac{x^2-2x}{x-2}\right)\\ \\ \end{align*}

Direct substitution of $x=2$ gives the indeterminate form $\frac{0}{0}$ . Therefore, we proceed by factoring the numerator.

\begin{align*} & =\lim\limits_{x\to 2}\left(\displaystyle \frac{x\left(x-2\right)}{x-2}\right) \\ \\ & =\lim\limits_{x\to 2}\left(x\right) \\ \\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}