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Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 22|

If f\left(x\right)=x^2-2x+3, find

\lim\limits_{x\to 0}\left(\frac{f\left(x+2\right)-f\left(2\right)}{x}\right)

SOLUTION:

\lim\limits_{x\to 0}\left(\frac{f\left(x+2\right)-f\left(2\right)}{x}\right)=\lim\limits_{x\to 0}\left(\frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(2^2-2\cdot 2+3\right)}{x}\right)

=\lim\limits_{x\to 0}\left(\frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(3\right)}{x}\right)

=\lim\limits_{x\to 0}\left(\frac{\left(\left(x+2\right)^2-2\left(x+2\right)\right)}{x}\right)

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

=\lim\limits_{x\to 0}\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