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
ln(x+h)-lnx = ln[1 + (h/x)=
-> h/x for x ->0
Divide that by h and you get 1/x. The limit as x->0 is infinity
I don't see how to use
limit of (e^h - 1)/h = 1 as h approaches 0 to solve this
e^h -1 -> h + h^2/2! + ... as h-> 0
To use the rule that says limit of (e^h - 1)/h = 1 as h approaches 0, we need to rewrite the expression [ln(x+h) - ln(x)]/h in a form that allows us to apply the rule.
First, let's simplify the expression by using the logarithmic property ln(a) - ln(b) = ln(a/b):
[ln(x+h) - ln(x)]/h = ln((x+h)/x)/h
Next, we can rewrite ln((x+h)/x) as ln(1 + h/x).
So now we have:
ln(1 + h/x)/h
To apply the limit rule, we need to rewrite the expression in the form (e^h - 1)/h.
Let's multiply the expression by x/x:
ln(1 + h/x)/h * (x/x)
This gives us:
x*ln(1 + h/x)/hx
Now, notice that the expression in the numerator, ln(1 + h/x), has the same form as (e^h - 1)/h. Therefore, we can use the limit rule and substitute it with 1 as h approaches 0.
The expression becomes:
x * (1/x) = 1
So, the limit of [ln(x+h) - ln(x)]/h as h approaches 0 is indeed 1/x, given that x>0.