How to evaluate the limit #[e^x-(1+x)]/x^3# as x approaches 0+?

#[e^x-(1+x)]/x^3#

1 Answer
VNVDVI
Apr 4, 2018

#lim_(x->0^+)(e^-x-(1+x))/x^3=oo#

Explanation:

Plugging in zero right away yields

#lim_(x->0^+)(e^x-(1+x))/x^3=(e^0-1)/0^3=0/0#

This is an indeterminate form and doesn't really tell us much; however, since we have an indeterminate form, we can use l'Hospital's Rule, which states that if

#lim_(x->a)f(x)/g(x)# yields an indeterminate form, then

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

So, we can differentiate the numerator and denominator of the limit individually, and see if that helps:

#f(x)=e^x-(1+x)=e^x-1-x#

#f'(x)=e^x-1#

#g(x)=x^3#

#g'(x)=3x^2#

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

We get an indeterminate form yet again. But we can apply l'Hospital's Rule as many times as necessary, so long as we end up with indeterminate forms each time.

#lim_(x->0^+)(e^x-1)/(3x^2)=lim_(x->0^+)e^x/(6x)=1/0=oo#

Thus,

#lim_(x->0^+)(e^x-(1+x))/x^3=oo#

This is a reasonable answer, as originally, we have an exponential in the numerator and a polynomial in the denominator, and exponentials grow far faster than polynomials, meaning the end result should be infinity.

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