# Evaluate the limit? lim_(x->0) (e^x-1-x-1/2x^2)/x^3

Jan 4, 2018

${\lim}_{x \to 0} \frac{{e}^{x} - 1 - x - \frac{1}{2} {x}^{2}}{x} ^ 3 = \frac{1}{6}$

#### Explanation:

We have that:

${\lim}_{x \to 0} \frac{{e}^{x} - 1 - x - \frac{1}{2} {x}^{2}}{x} ^ 3 = \frac{{e}^{0} - 1 - 0 - 0}{0} \to \frac{0}{0}$

So obviously we have an indeterminate form of $\frac{0}{0}$; this means we may use L'Hôpital's rule which states:

if ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) \to \frac{0}{0} \mathmr{and} \frac{\infty}{\infty}$ then:

${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

So taking the derivative of the numerator:

$\frac{d}{\mathrm{dx}} \left\{{e}^{x} - 1 - x - \frac{1}{2} {x}^{2}\right\} = {e}^{x} - 1 - x$

And the derivative of the numerator:

$\frac{d}{\mathrm{dx}} \left\{{x}^{3}\right\} = 3 {x}^{2}$ so:

${\lim}_{x \to 0} \frac{{e}^{x} - 1 - x - \frac{1}{2} {x}^{2}}{x} ^ 3 = {\lim}_{x \to 0} \frac{{e}^{x} - 1 - x}{3 {x}^{2}} \to \frac{0}{0}$

We have ran into an indeterminate form again, so we will apply L'Hôpital's rule a second time and differentiate again:

${\lim}_{x \to 0} \frac{{e}^{x} - 1 - x}{3 {x}^{2}} = {\lim}_{x \to 0} \frac{{e}^{x} - 1}{6 x} \to \frac{0}{0}$

Again we have ran into an indeterminate form so we need to keep going:

${\lim}_{x \to 0} \frac{{e}^{x} - 1}{6 x} = {\lim}_{x \to 0} \frac{{e}^{x}}{6} = \frac{1}{6}$

Therefore:

${\lim}_{x \to 0} \frac{{e}^{x} - 1 - x - \frac{1}{2} {x}^{2}}{x} ^ 3 = \frac{1}{6}$

So we finally got there in the end. A quick plot of the graph confirms our results:

As can be seen the line intersects the axis at 1/6.