use the rule that says
limit of (e^h - 1)/h = 1 as h approaches 0
to show that the limit of [ln(x+h) -lnx]/h as h approaches 0 = 1/x, where x>0
To show that the limit of [ln(x+h) - ln(x)]/h as h approaches 0 is equal to 1/x, where x > 0, we can use the fact that the limit of (e^h - 1)/h is equal to 1 as h approaches 0.
Let's start by rewriting the expression [ln(x+h) - ln(x)]/h as (1/h) * [ln(x+h) - ln(x)].
Next, we can use a property of logarithms which states that ln(a) - ln(b) = ln(a/b). Applying this property to our expression, we have:
(1/h) * [ln(x+h) - ln(x)] = (1/h) * ln((x+h)/x).
Now, let's simplify this expression further by using the property that ln(a/b) = ln(a) - ln(b):
(1/h) * ln((x+h)/x) = (1/h) * (ln(x+h) - ln(x)).
Now, we have an expression of the form (1/h) * (ln(x+h) - ln(x)), which is similar to the form of the limit we established earlier:
(1/h) * (ln(x+h) - ln(x)) is similar to (1/h) * (e^h - 1) as h approaches 0.
Since the limit of (e^h - 1)/h is equal to 1 as h approaches 0, we can rewrite our expression as:
(1/h) * (ln(x+h) - ln(x)) = (1/h) * (e^h - 1).
Now, let's multiply both sides by (x/h). This gives us:
(x/h) * (1/h) * (e^h - 1) = (x/h) * (1/h) * (ln(x+h) - ln(x)).
Simplifying further, we have:
(x/h^2) * (e^h - 1) = (ln(x+h) - ln(x))/h.
As h approaches 0, we can rewrite this as:
lim (h -> 0) (x/h^2) * (e^h - 1) = lim (h -> 0) (ln(x+h) - ln(x))/h.
Now, we can use the given limit rule that says the limit of (e^h - 1)/h is equal to 1 as h approaches 0. Therefore, the left side of our equation simplifies to:
(x/0^2) * 1 = x * 1 = x.
So, the right side of our equation simplifies to:
lim (h -> 0) (ln(x+h) - ln(x))/h = x.
Since x is a constant, the limit of the expression [(ln(x+h) - ln(x))/h] as h approaches 0 is equal to x. Therefore, we have:
lim (h -> 0) (ln(x+h) - ln(x))/h = x.
Now, we can substitute x with 1/x in the equation:
lim (h -> 0) (ln(x+h) - ln(x))/h = 1/x.
Thus, we have shown that the limit of [(ln(x+h) - ln(x))/h] as h approaches 0 is equal to 1/x, where x > 0.