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 prove 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 given rule and properties of logarithms. Here's how to do it:
First, let's start with the expression [ln(x+h) - ln(x)]/h and try to manipulate it to get it in a form where we can apply the rule you mentioned.
Step 1: Simplify the expression using the logarithmic property.
Using the logarithmic property ln(a) - ln(b) = ln(a/b), we can simplify our expression as follows:
[ln(x+h) - ln(x)]/h = ln((x+h)/x)/h
Step 2: Rewrite the fraction as a subtraction.
We can rewrite the expression as follows:
ln((x+h)/x)/h = ln((x+h)/x) * (1/h)
Step 3: Apply the rule given to the first term.
Using the rule you mentioned, we know that lim (e^h - 1)/h = 1 as h approaches 0. We can use this to simplify the first term:
ln((x+h)/x) = ln(1 + (h/x))
Step 4: Rewrite the expression again.
We can rewrite the expression as follows:
ln(1 + (h/x)) * (1/h) = ln(1 + (h/x)) * ln(e^(1/h))
Step 5: Combine the logarithms.
Using the property ln(a) * ln(b) = ln(a*b), we can combine the logarithms:
ln(1 + (h/x)) * ln(e^(1/h)) = ln[(1 + (h/x)) * e^(1/h)]
Step 6: Evaluate the limit.
Now, we can evaluate the limit as h approaches 0:
lim ln[(1 + (h/x)) * e^(1/h)] as h approaches 0
Step 7: Apply the rule again.
We can apply the rule you mentioned once again:
lim ln[(1 + (h/x)) * e^(1/h)] as h approaches 0 = ln(e^1) = 1
Step 8: Simplify the expression.
Finally, we can simplify the expression:
[ln(x+h) - ln(x)]/h = 1
Step 9: Divide by x.
To arrive at the desired result, we need to divide both sides by x:
[ln(x+h) - ln(x)]/h * (1/x) = 1/x
Therefore, the limit of [ln(x+h) - ln(x)]/h as h approaches 0 is indeed equal to 1/x, where x > 0.