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

2 Answers

6 votes

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

User Random Human
by
8.8k points
4 votes

Answer:

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

User Kalila
by
7.7k points

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