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PROBLEM:
If \displaystyle f\left(x\right)=x^2-2x+3, find \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(x+2\right)-f\left(2\right)}{x}\right).
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\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)\\
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& =\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)\\
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&=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)\right)}{x}\right)\\
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\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}\\
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& =\lim\limits_{x\to 0}\frac{\left(x+2\right)\left(x\right)}{x} \\
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& =\lim\limits_{x\to 0}\left(x+2\right) \\
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& =0+2 \\
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& =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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