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)... The notes for this section suck big time, so I really don't know where to begin with this. Thanks!
To verify the limit and understand how it relates to ƒ′(x) = ln e • ƒ(x) = ƒ(x), we need to break down the problem step by step.
Step 1: Start with the given expression:
lim h>0 (e^(x+h) - e^x) / h = e^x
Step 2: Expand the expression using the laws of exponents:
lim h>0 (e^x * e^h - e^x) / h = e^x
Step 3: Distribute e^x within the numerator:
lim h>0 (e^x * (e^h - 1)) / h = e^x
Step 4: Simplify further by factoring out e^x:
e^x * lim h>0 (e^h - 1) / h = e^x
Step 5: Now, focus on the term within the limit lim h>0 (e^h - 1) / h.
Step 6: Apply L'Hôpital's Rule to find the limit of the term:
lim h>0 [(d/dh(e^h - 1)) / (d/dh(h))] = lim h>0 (e^h) / 1 = 1
Step 7: Substitute the result of the limit back into the original expression:
e^x * 1 = e^x
Step 8: Since the limit of (e^x+h - e^x) / h as h approaches 0 is equal to e^x, we can conclude that ƒ′(x) = ln e • ƒ(x) = ƒ(x). Here's how it relates:
- The derivative of a function ƒ(x) is denoted by ƒ′(x).
- In this case, ƒ(x) = e^x.
- The derivative of e^x is e^x itself, which can be seen from the limit calculation we performed.
- Therefore, ƒ′(x) (the derivative of e^x) is equal to ln e (which is 1) multiplied by ƒ(x) (which is e^x), making it equal to ƒ(x).
This demonstrates the relationship between the derivative of e^x and the function itself.