Question

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

Answer 1

`lim_(x->0) (e^x-1-x-1/2x^2)/x^3=1/6`

Explanation:

We have that:

`lim_(x->0) (e^x-1-x-1/2x^2)/x^3=(e^0-1-0-0)/0-> 0/0`

So obviously we have an indeterminate form of `0/0`; this means we may use L'Hôpital's rule which states:

if `lim_(x->a) f(x)/g(x)-> 0/0 or oo/oo` then:

`lim_(x->a) f(x)/g(x)=lim_(x->a) (f'(x))/(g'(x))`

So taking the derivative of the numerator:

`d/dx{e^x-1-x-1/2x^2}=e^x-1-x`

And the derivative of the numerator:

`d/dx{x^3}=3x^2` so:

`lim_(x->0) (e^x-1-x-1/2x^2)/x^3=lim_(x->0)(e^x-1-x)/(3x^2)-> 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->0)(e^x-1-x)/(3x^2)=lim_(x->0)(e^x-1)/(6x)-> 0/0`

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

`lim_(x->0)(e^x-1)/(6x)=lim_(x->0)(e^x)/(6)=1/6`

Therefore:

`lim_(x->0) (e^x-1-x-1/2x^2)/x^3=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.