Question

Lim (c{x^2-2x-3}{x-3}" align="absmiddle" class="latex-formula">) as x tends to 3

Answer 1

The roots of the nominator are:

$\frac{ 2 ( + - ) \sqrt{4 + 12}}{2} = \frac{ 2 ( + - ) 4}{2} = 3 \: and \: - 1$

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:

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

Answer 2

Answer:

4

Step-by-step 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