Question

evaluate
$lim \: \: \: \: \: \: \: \frac{ {x}^{2 } - x - 6}{ {x}^{3} - 3 {x}^{2} + x - 3} \\ x - 3$

Answer 1

Hey there,

Numerator = [ x² - x - 6 ] = ( x - 3 )( x + 2 )

Denominator = [ x³ - 3x² + x - 3 ] = ( x - 3 )( x² + 1 )

=> The limit can now be written as :

[tex] \lim_{x \to 3} \frac{( x - 3 )( x + 2 )}{( x - 3 )( x^2 + 1 )} \\ \\ = \lim_{n \to \infty} \frac{x + 2}{x^2 + 1} [/tex]

Since this is no more an indeterminate form, we can substitute the value as :

→ The limiting value = ( 3 + 2 ) / ( 3² + 1 ) = ( 1 / 2 )

Answer 2

heya...!!!!



$lim \: \: \: \: \: \: \: \frac{ {x}^{2 } - x - 6}{ {x}^{3} - 3 {x}^{2} + x - 3} \\ x - 3 \\ \\ = > lim \: \: \: \: \: \: \: \frac{ {x}^{2 } - 3x + 2x - 6}{ { {x}^{2} (x - 3) + 1(x - 3)}} \\ x - 3 \\ \\ = > lim \frac{x(x - 3) + 2(x - 3)}{( {x}^{2} + 1)(x - 3) } \\ x - > 3 \\ \\ = > lim \frac{(x - 3)(x + 2)}{ ({x}^{2} +1 )(x - 3) } \\ x - > 3 \\ \\ = > lim \frac{(x + 2)}{( {x}^{2} + 1) } \\ x - > 3 \\ \\ = > lim \frac{3 + 2}{9 + 1} \\ x - > 3 \\ \\ = > \frac{5}{10} = \frac{1}{2}$