Suppose you're asked to find [math]\lim_{x\to1}\frac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)x}[/math] or [math]\lim_{x\to2}\frac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)x}[/math]. This is a tricky limit to take because:
both of these limits are of the indeterminate form [math]\frac{0}{0}[/math], meaning direct substitution will not work.
One way to evaluate these limits is to graph the rational function [math]y=\frac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)x}[/math] (or perform the analogous algebraic technique to determine the location of the holes in question).
[br][math]\lim_{x\to1}\frac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)x}=[/math]
[br][math]\lim_{x\to2}\frac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)x}=[/math]
But there is another approach too. Experiment with the applet below.
This applet illustrates an application of:
In the applet above [math]\frac{n\left(a\right)}{d\left(a\right)}[/math] and [math]\frac{n'\left(a\right)}{d'\left(a\right)}[/math] are only approximately equal near [math]a=1[/math] and [math]a=2[/math], but not otherwise. This serves as a reminder that:
L'Hopital's Rule only applies to limits of the indeterminate forms [math]\frac{0}{0}[/math] and [math]\frac{\pm\infty}{\pm\infty}[/math].