What is the limit of the following equation?

Limit as x approaches infinity of ((e^x) - ln(x)) / (x^3)

Since both numerator and denominator approach infinity as x→∞, we can apply l'Hôpital's rule:

(ex-(1/x))/3x²
Rewrite as a sum:
ex/3x² -(1/x)/3x²
the second term goes to zero as x→∞.
Apply the rule again to the first term:
ex/6x;
Apply the rule one last time:
ex/6; which goes to ∞ as x→&infin.
So the limit is ∞.

To find the limit of the given equation as x approaches infinity, you can use L'Hôpital's Rule, which states that if the limit of the ratio of two functions exists in an indeterminate form (such as 0/0 or ∞/∞), then the limit of their derivatives will also exist and have the same value.

Here's how you can apply L'Hôpital's Rule to this problem:

1. Take the derivative of the numerator and the denominator separately.
The derivative of e^x with respect to x is e^x.
The derivative of ln(x) with respect to x is 1/x.
The derivative of x^3 with respect to x is 3x^2.

2. Now rewrite the equation with the derivatives:
Limit as x approaches infinity of (e^x - ln(x)) / (x^3)
is equal to the limit as x approaches infinity of (e^x)/(3x^2).

3. Apply L'Hôpital's Rule again to the new equation:
Take the derivative of the numerator and the denominator separately.
The derivative of e^x with respect to x is e^x.
The derivative of 3x^2 with respect to x is 6x.

4. Now rewrite the equation with the derivatives:
Limit as x approaches infinity of (e^x)/(3x^2)
is equal to the limit as x approaches infinity of (e^x)/(6x).

5. Since e^x grows faster than any power of x as x approaches infinity, the limit of (e^x)/(6x) as x approaches infinity will be infinity.

Therefore, the limit of the given equation as x approaches infinity is infinity.