Answer
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When $x=2.8$ ; $\frac{x^{2}-9}{x-3}$=$\frac{7.84-9}{2.8-3}$=5.8
When $x=2.9$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.41-9}{2.9-3}$=5.9
When $x=2.95$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.7025-9}{2.95-3}$=5.95
When $x=2.99$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.9401-9}{2.99-3}$=5.99
When $x=2.999$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.994001-9}{2.999-3}$=5.999
When $x=3.2 ; $$ \frac{x^{2}-9}{x-3}$=$\frac{10.24-9}{2.8-3}$=6.2
When $x=3.1$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.61-9}{2.9-3}$=6.1
When $x=3.05$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.3025-9}{2.95-3}$=6.05
When $x=3.01$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.0601-9}{2.99-3}$=6.01
When $x=3.001$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.006001-9}{2.999-3}$=6.001
Work Step by Step
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We have to consider what the rational expression approaches as $x$ gets "infinitely" closer to 3. As we can approach 3 from both higher and lower values we will try both of them (So we can determine definition of a limit):
When $x=2.8$ ; $\frac{x^{2}-9}{x-3}$=$\frac{7.84-9}{2.8-3}$=5.8
When $x=2.9$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.41-9}{2.9-3}$=5.9
When $x=2.95$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.7025-9}{2.95-3}$=5.95
When $x=2.99$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.9401-9}{2.99-3}$=5.99
When $x=2.999$ ; $\frac{x^{2}-9}{x-3}$=$\frac{8.994001-9}{2.999-3}$=5.999
When $x=3.2 ; $$ \frac{x^{2}-9}{x-3}$=$\frac{10.24-9}{2.8-3}$=6.2
When $x=3.1$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.61-9}{2.9-3}$=6.1
When $x=3.05$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.3025-9}{2.95-3}$=6.05
When $x=3.01$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.0601-9}{2.99-3}$=6.01
When $x=3.001$ ; $\frac{x^{2}-9}{x-3}$=$\frac{9.006001-9}{2.999-3}$=6.001
(*)
If we continue to approach $x$ "infinitely" close to 3, the definition of the expression will get "infinitely" close to 6.
To get clear definition of why this happens, let's factorize the numerator of this expression :
$\frac{x^{2}-9}{x-3}$=$\frac{(x-3)(x+3)}{(x-3)}$ = $x+3$
So now it's clear, that if $x=3$ expression gets 6. But as the base of this expression forbids us to define $x$ as 3 we can't grant $x$ to be .
(Limit of the expression as $x$ approaches 3 is 6)
