Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.lim x→0 e5x − 1 − 5xx2

To find the limit of the given function as x approaches 0, we can use l'Hospital's Rule. l'Hospital's Rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives, provided the limit of the original functions is in the form 0/0 or ∞/∞.

The given function is (e^(5x) - 1 - 5x) / x^2. As x approaches 0, both the numerator and the denominator approach 0, so we can apply l'Hospital's Rule.

First, we need to find the derivative of the numerator and the derivative of the denominator.

The derivative of e^(5x) - 1 - 5x with respect to x is 5e^(5x) - 5.

The derivative of x^2 with respect to x is 2x.

So, according to l'Hospital's Rule, the limit of the original function as x approaches 0 is equal to the limit of (5e^(5x) - 5) / 2x as x approaches 0.

However, this is still in the form 0/0, so we can apply l'Hospital's Rule again.

The derivative of 5e^(5x) - 5 with respect to x is 25e^(5x).

The derivative of 2x with respect to x is 2.

So, the limit of the original function as x approaches 0 is equal to the limit of 25e^(5x) / 2 as x approaches 0.

As x approaches 0, e^(5x) approaches 1, so the limit is 25/2, or 12.5.