Question

Lim (
`(x^2-2x-3)/(x-3)`) as x tends to 3

Answer 1

The roots of the nominator are:

`( 2 ( + - ) √(4 + 12))/(2) = ( 2 ( + - ) 4)/(2) = 3 \: and \: - 1`

Which means that the nominator can be written as:

`{x}^(2) - 2x - 3 = (x - 3)(x + 1)`

Thus we compute the limit as shown below:

`lim( \frac{ {x}^(2) - 2x - 3}{x - 3})(x - > 3) = lim( ((x - 3)(x + 1))/(x - 3))(x - > 3) = lim( x + 1)(x - > 3) = 4`

Answer 2

4

Explanation:

If you replace x with 3 in the expression you will get an indefenite form (0/0)

So there two methods.

The easiest one is applying The hospital rule

Derivate the expression firsr but separatly.

● (x^2 -2x -3)' = 2x - 2

● (x-3)' = 1

When x tends to 3

Lim(x^2-2x-3/x-3) = lim (2x-2/1) = 6-2 = 4