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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