Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 21|

If $f\left(x\right)=x^2-2x+3$ , find

$\lim\limits_{x\to 2}\left(\frac{f\left(x\right)-f\left(2\right)}{x-2}\right)$

SOLUTION:

$\lim\limits_{x\to 2}\left(\frac{f\left(x\right)-f\left(2\right)}{x-2}\right)=\lim\limits_{x\to 2}\left(\frac{\left(x^2-2x+3\right)-\left(2^2-2\cdot 2+3\right)}{x-2}\right)$

$=\lim\limits_{x\to 2}\left(\frac{\left(x^2-2x+3\right)-3}{x-2}\right)$

$=\lim\limits_{x\to 2}\left(\frac{x^2-2x}{x-2}\right)$

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

$=\lim\limits_{x\to 2}\left(\frac{x\left(x-2\right)}{x-2}\right)$

$=\lim\limits_{x\to 2}\left(x\right)$

$=2$

If , find

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