lim (1+x)^1/x. Give an exact answer.
x->0
This reads:
The limits as x approaches zero is (1 plus x) to the 1 divided by x.
The log of each term is
(1/x) ln (1 + x) = ln (1+x)/x
Using L'Hopital's rule for the limit of
f(x)/g(x), the limit if the log is
lim f'/g' = lim [1/(1+x)]= 1
and the antilog of that is e. 2.71828...
To find the exact answer to the limit, lim (1+x)^(1/x) as x approaches 0, you can first take the natural logarithm of both sides of the expression.
ln(lim (1+x)^(1/x)) = ln((1+x)^(1/x))
Then, you can use the property that the logarithm of a limit is equal to the limit of the logarithm:
lim ln((1+x)^(1/x))
Now, you can simplify the expression by using the property of logarithms that states ln(a^b) = b*ln(a):
lim (1/x) * ln(1+x)
Next, you can use L'Hôpital's rule, which is a method for evaluating limits of indeterminate forms, where both the numerator and denominator approach 0 or infinity. In this case, the limit of 1/x as x approaches 0 is infinity:
lim (1/x) * ln(1+x) = lim (1/(1+x))
Evaluating this limit as x approaches 0, you get:
lim (1/(1+0)) = 1
Therefore, the value of the natural logarithm of the original limit is equal to 1. To find the original limit, you can take the antilog of 1, which gives you:
e^1 = e
Hence, the exact answer to the limit is e, approximately equal to 2.71828.