Question

`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

Answer 1

`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`