Answer
$${\text{ The limit does not exist}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{1 + 2x - {{\left( {1 + x} \right)}^{1/2}}}}{{{x^3}}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {1 + 2x - {{\left( {1 + x} \right)}^{1/2}}} \right) = 1 + 2\left( 0 \right) - {\left( {0 + x} \right)^{1/2}} = 0 \cr
& \mathop {\lim }\limits_{x \to 0} {x^3} = {0^3} = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{}} \cr
& {\text{for }}1 + 2x - {\left( {1 + x} \right)^{1/2}} \to {D_x}\left( {1 + 2x - {{\left( {1 + x} \right)}^{1/2}}} \right) = 2 - \frac{1}{2}{\left( {1 + x} \right)^{ - 1/2}} \cr
& {\text{for }}{x^3} \to {D_x}\left( {{x^3}} \right) = 3{x^2} \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{1 + 2x - {{\left( {1 + x} \right)}^{1/2}}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2 - \frac{1}{2}{{\left( {1 + x} \right)}^{ - 1/2}}}}{{3{x^2}}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{2 - \frac{1}{2}{{\left( {1 + 0} \right)}^{ - 1/2}}}}{{3{{\left( 0 \right)}^2}}} = \frac{{3/2}}{0} \cr
& {\text{Since the limits do not lead to the indeterminate form }}\frac{0}{0}{\text{ or }}\frac{{ \pm \infty }}{{ \pm \infty }} \cr
& {\text{we cannot apply the l'Hospital's rule}}. \cr
& {\text{By l'Hospital' rule}}{\text{, the limit does not exist}} \cr} $$
