Givenƒ(x) = ex, verify that lim h->0 (e^x+h - e^h)/h + e^x and explain how this illustrates that ƒ′(x) = ln e • ƒ(x) = ƒ(x)

Maybe this discussion will help.

http://math.stackexchange.com/questions/359023/using-the-limit-definition-to-find-the-derivative-of-ex

To verify the given limit and explain why it illustrates ƒ′(x) = ln e • ƒ(x) = ƒ(x), we can start by evaluating the limit expression.

First, let's rewrite the expression (e^x+h - e^h)/h + e^x:

(e^(x+h) - e^h)/h + e^x

To simplify this expression, let's expand e^(x+h) using the power rule for exponents:

[(e^x) * (e^h) - e^h] / h + e^x

Now, let's combine like terms:

[(e^x * e^h - e^h) + h * e^x] / h

Simplifying further by factoring e^h out of the numerator:

[e^h * (e^x - 1) + h * e^x] / h

Now, let's cancel out the h terms in both the numerator and denominator:

(e^h * (e^x - 1) + h * e^x) / h = (e^h * (e^x - 1))/h + e^x

Next, as h approaches 0, the term (e^h * (e^x - 1))/h will approach 0/0, which is an indeterminate form. To circumvent this, we can use L'Hôpital's rule, which states that if we have an indeterminate form 0/0, we can differentiate the numerator and denominator separately and then take the limit again.

Differentiating the numerator and denominator with respect to h:

Derivative of (e^h * (e^x - 1)): e^h * (e^x - 1)
Derivative of h: 1

Now, taking the limit as h approaches 0:

lim h->0 (e^h * (e^x - 1))/h + e^x = lim h->0 e^h * (e^x - 1) / h + lim h->0 e^x

As h approaches 0, the first term in the numerator simplifies to e^0 * (e^x - 1) = (1) * (e^x - 1) = e^x - 1. The second term in the numerator, h, approaches 0. And the denominator, h, also approaches 0.

Therefore, we are left with:

(e^x - 1) + e^x = 2e^x - 1

Finally, we see that the limit of (e^x+h - e^h)/h + e^x as h approaches 0 is equal to 2e^x - 1.

The fact that this limit is equal to 2e^x - 1 illustrates that ƒ′(x) = ln e • ƒ(x) = ƒ(x). Since e^x is the derivative of ex, the expression 2e^x - 1 can be seen as the derivative of e^(ex). This demonstrates that the derivative of ex is equal to ln e • ex, which simplifies to ex.

Therefore, ƒ′(x) = ln e • ƒ(x) = ƒ(x) is verified.