If |x|< 1/2 then the polynomial $1 + \frac{x}{2}-\frac{x^2}{8}$ aproximates the value of $\sqrt{1+x}$ with an error minor of $\frac{1}{2}|x^3|$

Question

Demonstrate that if $|x|< 1/2$ then the polynomial $1 + \frac{x}{2}-\frac{x^2}{8}$ aproximates the value of $\sqrt{1+x}$ with an error minor of $\frac{1}{2}|x^3|$.

I've done the MacLaurin polinomial of $\sqrt{1+x}$ until degree $2$ that is $1 + \frac{x}{2}-\frac{x^2}{8}$. I know how to calculate the error depending on the grade of the polynomial but not how to calculate it depending on the value of $x$. I also tried to use the Lagrange reminder.