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

(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.

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

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

So helpful, thank you!!

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.

No worries! I can help you with that.

To evaluate the limit as x approaches zero of (e^(6/x)-6x)^(x/2), we can indeed apply L'Hôpital's Rule. It can be a bit tricky to figure out the best way to apply it here, but we will break it down step by step to make it easier to understand.

First, let's rewrite the given expression in a suitable form for applying the rule:

(e^(6/x) - 6x)^(x/2) = e^((6/x) * (x/2)) - 6x^(x/2)

Now let's focus on evaluating the limit of each term separately.

1. The first term: e^((6/x) * (x/2))
By properties of exponents, we can rewrite this as e^(3), since (6/x) * (x/2) simplifies to 3.
Therefore, the limit of this term as x approaches zero is e^3.

2. The second term: -6x^(x/2)
Here we have an indeterminate form (0 * infinity) since the base approaches zero (x^(x/2) goes to 0) and the exponent (x/2) approaches infinity.

To evaluate this term, we will use L'Hôpital's Rule on the natural log of the expression to get rid of the exponent:

ln(-6x^(x/2)) = ln(-6) + ln(x^(x/2))

Now, applying L'Hôpital's Rule on ln(x^(x/2)), we take the derivative of the numerator and denominator:

d/dx ln(x^(x/2)) = d/dx (x/2) * ln(x)
= (1/2) * ln(x) + (x/2) * (1/x)
= (1/2) * ln(x) + 1/2

Now let's evaluate the limit of this expression as x approaches zero:

lim(x->0) [(1/2) * ln(x) + 1/2] = (1/2) * ln(0) + 1/2
= (1/2) * (-∞) + 1/2
= -∞ + 1/2
= -∞

Overall, we have e^3 * (-∞) = -∞ as the limit of the second term.

Therefore, the final limit of the entire expression is -∞.

I hope this explanation helps! Let me know if you have any further questions.