Solución

Paso 1.

x+2−1x+1{{\sqrt {x + 2} - 1} \over {x + 1}} es de la forma 0/00/0 en el punto 11. Operando lim⁡x→11x+1−12x−1=lim⁡x→12−(x+1)2(x+1)x−1=lim⁡x→11−x2(x−1)(x+1)\mathop {\lim }\limits_{x \to 1} {{{1 \over {x + 1}} - {1 \over 2}} \over {x - 1}} = \mathop {\lim }\limits_{x \to 1} {{{{2 - \left( {x + 1} \right)} \over {2\left( {x + 1} \right)}}} \over {x - 1}} = \mathop {\lim }\limits_{x \to 1} {{1 - x} \over {2\left( {x - 1} \right)\left( {x + 1} \right)}}

Paso 2.

Simplificando =lim⁡x→1−12(x+1) = \mathop {\lim }\limits_{x \to 1} {{ - 1} \over {2\left( {x + 1} \right)}}

Paso 3.

Aplicando finalmente las propiedades de los límites

lim⁡x→1−12(x+1)=−14\mathop {\lim }\limits_{x \to 1} {{ - 1} \over {2\left( {x + 1} \right)}} = - {1 \over 4}