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

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1 Answer
Jan 4, 2018

#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:

Generated on Mathematica

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