#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?

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

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Answer 1 · 2018-03-27T21:54:55

#8#

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#