Answer
$\dfrac{-1}{32}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{\sqrt {1+x}-1-\dfrac{x}{2}}{4x^2}$
Recall the series for $\sqrt {1+x} =\Sigma_{k =0}^{\infty} \dbinom{1/2}{k}x^k$
Now, $L=\lim\limits_{x \to 0} \dfrac{\Sigma_{k =0}^{\infty}\dbinom{1/2}{k}x^k-1-\dfrac{x}{2}}{4x^2} \\=\lim\limits_{x \to 0} [\Sigma_{k =0}^{\infty}\dbinom{1/2}{k}\dfrac{x^{k-2}}{4}-\dfrac{1}{4x^2}-\dfrac{1}{8x}]\\=\lim\limits_{x \to 0} [(\dbinom{1/2}{0}\dfrac{x^{0-2}}{4}+\dbinom{1/2}{1}\dfrac{x^{1-2}}{4}+........)-\dfrac{1}{4x^2}-\dfrac{1}{8x}]\\=\lim\limits_{x \to 0} [(\dfrac{1}{4x^2}-\dfrac{1}{8x}......)-\dfrac{1}{4x^2}-\dfrac{1}{8x}]\\=\dfrac{-1}{32}+0+0+0+.......\\=\dfrac{-1}{32}$
