Find
a) lim (ln x)^1/x
x->infinity
b) lim (x-1)^lnx
x->1+
lim (lnx)^1/x
take the antilog
(e^x)^(1/x)=e^1=e
take the ln...
lim= 1
To find the limit of a function as x approaches a certain value, we need to determine the behavior of the function as x gets arbitrarily close to that value.
a) To find the limit of (ln x)^(1/x) as x approaches infinity, we need to understand the properties of logarithmic and exponential functions.
We can rewrite the given expression as e^(ln[(ln x)^(1/x)]) since e^ln(y) is just equal to y.
Next, we can use the properties of logarithms to rewrite the expression as e^[(1/x) * ln(ln x)].
Now, we can see that as x approaches infinity, both 1/x and ln(ln x) tend towards 0.
So, the term (1/x) * ln(ln x) approaches 0 * 0, which is an indeterminate form.
To evaluate the limit, we can use L'Hôpital's Rule, which states that if we have an indeterminate form of the type 0/0 or ∞/∞, we can take the derivative of both the numerator and denominator and then re-evaluate the limit.
Let's take the derivative of (1/x) * ln(ln x):
d/dx [(1/x) * ln(ln x)] = (1/x) * (1/ln x) * (1/x)
Next, we simplify the derivative:
= (1/x^2) * (1/ln x)
Now, let's rewrite the original limit with the derivative:
lim (x->infinity) [(1/x) * ln(ln x)] = lim (x->infinity) [(1/x^2) * (1/ln x)]
To evaluate this new limit, we can again use L'Hôpital's Rule and take the derivative of both the numerator and the denominator:
d/dx [(1/x^2) * (1/ln x)] = (1/x^2) * (-2/x^3) * (1/ln x) + (1/x^2) * (-1/x) * (1/x * (1/x))
Simplifying the derivative:
= (-2/x^5) * (1/ln x) + (-1/x^4) * (1/x^2)
Now, we can evaluate the limit of this new expression as x approaches infinity:
lim (x->infinity) [(-2/x^5) * (1/ln x) + (-1/x^4) * (1/x^2)] = 0 + 0 = 0
Therefore, the limit of (ln x)^(1/x) as x approaches infinity is 0.
b) To find the limit of (x-1)^(ln x) as x approaches 1+, we can use the concept of continuity.
Let's consider the natural logarithm function ln x. It is continuous for x > 0.
Since we are approaching 1+ from the positive side, ln x will be defined and continuous for the values of x near 1.
Now, let's examine the expression (x-1)^(ln x):
As x approaches 1+, the term (x-1) tends towards 0 and ln x is defined and continuous near 1.
For any positive value of ln x, (x-1)^(ln x) will also approach 0.
Therefore, the limit of (x-1)^(ln x) as x approaches 1+ is 0.