Givenƒ(x) = e^x, verify that
lim (e^x+h-e^h) / h = e^x
h->O =
and explain how this illustrates that ƒ′(x) = ln e • ƒ(x) = ƒ(x)
This looks like a Derek question. Please use a single name for your posts
I'm not Derek, I've posted before thanks. We're friends and we take the same class.
To verify the given limit and understand how it relates to ƒ′(x) = ln e • ƒ(x) = ƒ(x), let's work through the steps together.
First, let's apply the limit definition:
lim (e^x+h - e^h) / h
h->0
Now, we can simplify the expression. Since we have exponential functions, we can utilize the properties of exponents. Recall that e^a + b can be rewritten as e^a * e^b. Applying this property, we get:
lim (e^x * e^h - e^h) / h
h->0
Next, we can factor out e^h from the numerator:
lim (e^h * (e^x - 1)) / h
h->0
Now, we have (e^h * (e^x - 1)) in the numerator and h in the denominator. The goal is to simplify the expression further to prove that it equals e^x. To do so, let's focus on the term e^h * (e^x - 1). As h approaches 0, e^h approaches 1 because any positive number raised to power 0 is 1. Hence, as h->0, e^h approaches 1, and (e^x - 1) remains unchanged. So, we can rewrite the expression as:
lim (1 * (e^x - 1)) / h
h->0
Now, let's simplify further:
lim (e^x - 1) / h
h->0
Finally, as h approaches 0, the expression (e^x - 1) / h represents the derivative of e^x with respect to x, which is ƒ′(x). Therefore, the limit can be represented as:
lim ƒ′(x)
h->0
And since ƒ′(x) = ƒ(x) (the derivative of e^x is e^x), we can substitute it in:
lim ƒ(x)
h->0
This proves that the limit (e^x+h - e^h) / h as h approaches 0 is indeed equal to e^x, and it also illustrates that ƒ′(x) = ln e • ƒ(x) = ƒ(x), as the derivative of e^x is itself.