limx->0 ((e^(4x)-1-4x)/(x^2)) using l'hopital's rule limit as x approaches 0 ((e^(4x)-1-4x)/(x^2) using l'hopital's rule?

limx->0 ((e^(4x)-1-4x)/(x^2)) using l'hopital's rule

1 Answer
Mar 27, 2018

8

Explanation:

Plugging in 0 right away yields

lim_(x->0)(e^(4x)-1-4x)/x^2=(e^0-1)/0=0/0, an indeterminate form.

Now, l'Hospital's Rule tells us if we have a limit in the form

lim_(x->a)f(x)/g(x) and the limit yields an indeterminate form (such as 0/0, (+-oo)/(+-oo), then lim_(x->a)f(x)/g(x)=lim_(x->a)(f'(x))/(g'(x))

Here, we see f(x)=e^(4x)-1-4x

Then,

f'(x)=4e^(4x)-4

g(x)=x^2

g'(x)=2x

So,

lim_(x->0)(e^(4x)-1-4x)/x^2=lim_(x->0)(4e^(4x)-4)/(2x)=(4-4)/0=0/0

Another indeterminate form. Fortunately, the rule can be applied as many times as necessary, so long as you're getting indeterminate forms.

So, apply again by differentiating numerator and denominator:

lim_(x->0)(4e^(4x)-4)/(2x)=lim_(x->0)(16e^(4x))/2=16/2=8