Show that the limit as h approaches 0 of (e^h-1 ) all over h = ln e = 1 using at least two numerical examples.
To verify the limit as h approaches 0 of the given expression, we can evaluate the expression for small values of h using at least two numerical examples and observe the trend.
Let's compute the expression for h = 0.1 and h = 0.01.
For h = 0.1:
(e^h - 1) / h = (e^0.1 - 1) / 0.1
Using a calculator, we find that e^0.1 is approximately 1.1052.
Substituting this value into the expression, we have:
(1.1052 - 1) / 0.1 ≈ 0.1052 / 0.1 ≈ 1.052
For h = 0.01:
(e^h - 1) / h = (e^0.01 - 1) / 0.01
Again, using a calculator, we find that e^0.01 is approximately 1.0101.
Substituting this value into the expression, we have:
(1.0101 - 1) / 0.01 ≈ 0.0101 / 0.01 ≈ 1.01
As we can see, as h approaches 0, the evaluated values of the expression are getting closer to 1. In both numerical examples, we obtained values very close to 1.
However, to establish the limit more rigorously, we can use calculus. Taking the limit as h approaches 0, we can make use of the derivative of the natural logarithm function:
lim(h->0) [(e^h - 1) / h] = lim(h->0) [d/dh (ln(e^h))]
Applying the derivative:
lim(h->0) [e^h / e^h] = lim(h->0) [1] = 1
Therefore, the limit of (e^h - 1) / h as h approaches 0 is indeed 1, which can be verified using both numerical examples and calculus.