Question

Evaluate the limit ( 1 / (x² + x - 2) - x/(x³-1) ) when x tends to 1.​

Answer 1

Step-by-step explanation:

We have,

$\tt{ \blue{ \lim_{x \to1} \bigg( \dfrac{1}{ {x}^{2} + x - 2} - \dfrac{x}{ {x}^{3} - 1 } \bigg)}}$

$\sf{ = \lim_{x \to1} \bigg( \dfrac{1}{ {x}^{2} +2 x - x - 2} - \dfrac{x}{ (x - 1)( {x}^{2} + x + 1) } \bigg)}$

$\sf{ = \lim_{x \to1} \bigg \{\dfrac{1}{ x(x +2)-( x + 2)} - \dfrac{x}{ (x - 1)( {x}^{2} + x + 1) } \bigg \}}$

$\sf{ = \lim_{x \to1} \bigg \{\dfrac{1}{( x - 1)(x +2)} - \dfrac{x}{ (x - 1)( {x}^{2} + x + 1) } \bigg \}}$

$\sf{ = \lim_{x \to1} \: \: \dfrac{1}{x - 1} \bigg \{\dfrac{1}{x +2} - \dfrac{x}{ {x}^{2} + x + 1 } \bigg \}}$

$\sf{ = \lim_{x \to1} \: \: \dfrac{1}{x - 1} \bigg \{ \dfrac{ {x}^{2} + x + 1 - x(x + 2)}{ (x + 2) ({x}^{2} + x + 1 )} \bigg \}}$

$\sf{ = \lim_{x \to1} \: \: \dfrac{1}{x - 1} \bigg \{ \dfrac{ {x}^{2} + x + 1 - x^{2} - 2x}{ (x + 2) ({x}^{2} + x + 1 )} \bigg \}}$

$\sf{ = \lim_{x \to1} \: \: \dfrac{1}{x - 1} \bigg \{ \dfrac{ 1 - x}{ (x + 2) ({x}^{2} + x + 1 )} \bigg \}}$

$\sf{ = \lim_{x \to1} \: \: \dfrac{1}{x - 1} \bigg \{ \dfrac{ -( x - 1)}{ (x + 2) ({x}^{2} + x + 1 )} \bigg \}}$

$\sf{ = - \lim_{x \to1} \: \: \bigg \{ \dfrac{ 1}{ (x + 2) ({x}^{2} + x + 1 )} \bigg \}}$

$\sf{ = - \: \: \bigg \{ \dfrac{1}{ (1 + 2) ((1)^{2} + 1 + 1 )} \bigg \}}$

$\sf{ = - \: \: \dfrac{1}{ (3) (3)} }$

$\sf{ = - \: \: \dfrac{1}{9} }$