Find the limit as h approaches 0 for
ln(1+h)/h
use l'Hospital's rule. The limit is the same as
[1/(1+h)]/1 = 1
To find the limit as h approaches 0 for ln(1+h)/h, we can use L'Hospital's Rule.
L'Hospital's Rule states that if we have a limit in the form of 0/0 or ∞/∞, we can differentiate the numerator and denominator separately until we get a limit that is not in indeterminate form.
Let's apply L'Hospital's Rule to our limit:
lim(h→0) ln(1+h)/h
Since we have a limit in the form of 0/0, we can differentiate the numerator and denominator separately:
lim(h→0) ln(1+h) / h
= lim(h→0) (d/dh ln(1+h)) / (d/dh h)
To differentiate ln(1+h), we use the chain rule:
d/dh ln(1+h) = 1 / (1+h) * (d/dh (1+h))
= 1 / (1+h)
Differentiating h with respect to h is simply 1.
Now we can rewrite our limit using the differentiated forms:
lim(h→0) 1 / (1+h) / 1
= lim(h→0) 1 / (1+h)
Now we can plug in h=0 to find the limit:
lim(h→0) 1 / (1+h) = 1 / (1+0) = 1
Therefore, the limit as h approaches 0 for ln(1+h)/h is 1.
To find the limit as h approaches 0 for the expression ln(1+h)/h, we can use L'Hôpital's rule, which states that if the limit of f(x)/g(x) as x approaches a is of the form 0/0 or ∞/∞, then the limit is equal to the limit of the derivative of f(x) divided by the derivative of g(x) as x approaches a.
Let's apply L'Hôpital's rule to this problem:
Take the derivative of the numerator and denominator separately:
d/dh (ln(1+h)) = 1/(1+h)
d/dh (h) = 1
Now, find the limit as h approaches 0 for the derivative of the original expression:
lim(h->0) (1/(1+h))/1
Simplifying further:
lim(h->0) 1/(1+h)
Now, substitute h = 0 into the expression and evaluate the limit:
1/(1+0) = 1/1 = 1
Therefore, as h approaches 0, the limit of ln(1+h)/h is equal to 1.