Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 22

PROBLEM:

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SOLUTION:

\begin{align*} \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(x+2\right)-f\left(2\right)}{x}\right)&=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(2^2-2\cdot 2+3\right)}{x}\right)\\ \\ & =\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(3\right)}{x}\right)\\ \\ &=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)\right)}{x}\right)\\ \\ \end{align*}

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

\begin{align*} & =\displaystyle \lim\limits_{x\to 0}\displaystyle \frac{\left(x+2\right)\left(x+2-2\right)}{x}\\ \\ & =\lim\limits_{x\to 0}\frac{\left(x+2\right)\left(x\right)}{x} \\ \\ & =\lim\limits_{x\to 0}\left(x+2\right) \\ \\ & =0+2 \\ \\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}