Answer
$\dfrac{-1}{8}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{1+x-e^x}{4x^2}$
Recall the series for $e^x=\Sigma_{k =0}^{\infty} \dfrac{(x)^{k}}{k!}$
Now, $L=\lim\limits_{x \to 0} \dfrac{1+x-\Sigma_{k =0}^{\infty} \dfrac{(x)^{k}}{k!}}{4x^2} \\=\lim\limits_{x \to 0} ( \dfrac{1}{4x^2}+\dfrac{1}{4x}-\Sigma_{k =0}^{\infty} \dfrac{(x)^{k-2}}{4 k!}) \\=\lim\limits_{x \to 0} ( \dfrac{1}{4x^2}+\dfrac{1}{4x}-[ \dfrac{(x)^{0-2}}{4 (0!)}+\dfrac{(x)^{1-2}}{4 (1!)}+.....])\\=\lim\limits_{x \to 0} [\dfrac{-1}{8}-\dfrac{x}{(4)(3!)}-\dfrac{x^2}{(4)(4!)}......]\\=\dfrac{-1}{8}-\dfrac{x}{(4)(3!)}-\dfrac{x^2}{(4)(4!)}......\\=\dfrac{-1}{8}$
