$f(x) = \frac{x^{3}-1}{x-1}$

As the denominator contains x-1, so we get confused what to do when x=1. So to get an idea about the behavior of the graph near to x= 1. We will use the two set of values of x, one set of values from left side approaches to 1 and other set of values from the right side approaches to 1.

The graph of the given function is a parabola but there is a gap at (1,3) as shown in the above graph. From the graph and the table will understand that x can not be equal to 1 and the f(x) or y will be close to 3. So we will write this using limit notation as

$\lim_{x->1}f(x) = 3$

This can be read as

$f(x)=\frac{x-4}{x^{2}-3x-4}$

$ \lim_{x->4}\frac{x-4}{x^{2}-3x-4}$

x values from the left side = 3.9,3.99,3.999,4

x values from the right side = 4.001,4.01,4.1

We will plug in the above value in the given function.

x= 3.9

$f(3.9)=\frac{3.9-4}{3.9^{2}-3(3.9)-4} =0.2040 $

x=3.99

$f(3.99)=\frac{3.99-4}{3.99^{2}-3(3.99)-4} =0.200400 $

x=3.999

$f(3.999)=\frac{3.999-4}{3.999^{2}-3(3.999)-4} =0.20004000 $

x= 4

$f(4)=\frac{4-4}{4^{2}-3(4)-4} = undefined (no value)$

x= 4.001

$f(4.001)=\frac{4.001-4}{4.001^{2}-3(4.001)-4} = 1.99960$

x= 4.01

$f(4.01)=\frac{4.01-4}{4.01^{2}-3(4.01)-4} = 1.99601$

x=4.1

$f(4.1)=\frac{4.1-4}{4.1^{2}-3(4.1)-4} = 1.996078$

From the above, we can see that when we get closer to 4 the value gets closer to 0.20. So we can conclude that

$ \lim_{x->4}\frac{x-4}{x^{2}-3x-4} \approx 0.2$

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