Using l'hopital's rule, find the limit as x approaches zero of

Question

(e^(6/x)-6x)^(X/2)
I know l'hopital's rule, but this is seeming to be nightmare. I just don't seem to get anywhere.
I mistyped the first time I posted this question.

Answer 1

This is not an obvious one.
(I changed the uppercase X to lowercase x)

Lim x->0 (e^(6/x)-6x)^(x/2)
At first glance, as x->0, 6/x->∞ and 6x->0.
By comparison of the two terms, the -6x term can be removed to give:
Lim x->0 (e^(6/x))^(x/2)

By the use of the laws of exponent, the above expression reduces to:
Lim x->0 e^(6x/2x)

Apply d'Hôpital's rule to the exponent, we get
Lim x->0 e^(6/2)
=e^3

Answer 2

Thank you so much! I didn't think to remove the 6x.
So helpful, thank you!!

Answer 3

Here's what's weird: I graphed the equation and from the left the function approaches 1 as x approaches zero, and from the right it approaches e^3. Doesn't that mean the limit does not exist? The question, however, is multiple choice, with choices of infinity, 0, 3, 1, and e^3.