# Confusing log limit

## Homework Statement

$$\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x$$

## Homework Equations

Maclaurin series:
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ... + (-1)^{r+1} \frac{x^r}{r} + ...$$

## The Attempt at a Solution

We're considering vanishingly small $x$, so just taking the first term in the Maclaurin series the limit becomes:

$$\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x = \lim\limits_{x \to 0} x^x = \mathrm{undefined}$$

or so I thought until google tells me that $0^0 = 1$.

What's going on here? How can I evaluate the limit properly?

member 587159

## Homework Statement

$$\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x$$

## Homework Equations

Maclaurin series:
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ... + (-1)^{r+1} \frac{x^r}{r} + ...$$

## The Attempt at a Solution

We're considering vanishingly small $x$, so just taking the first term in the Maclaurin series the limit becomes:

$$\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x = \lim\limits_{x \to 0} x^x = \mathrm{undefined}$$

or so I thought until google tells me that $0^0 = 1$.

What's going on here? How can I evaluate the limit properly?
0^0 is undefined

RUber
Homework Helper
Because ##0^0## is undefined, you have to take the limit, as was given in the problem.
##\lim_{x\to 0} (\ln(1+x))^x ## can be looked at as a related rates sort of problem...is the log part going to zero faster then the exponent is taking it back out to 1?
In either case, the expected options for the solution should be 0 or 1.
Let's look at this as a sequence:
##\{x_n\} = [ \ln ( 1 + \frac{1}{2^n}) ]^ {\frac{1}{2^n} }##
##\lim_{x\to 0} (\ln(1+x))^x = \lim_{n\to \infty} x_n##
For n = 1, ##x_n= \ln(2)##
Now check the ratio ##\frac{x_n}{x_{n+1}}##. If this is >1, then the sequence is shrinking...if it is <1, then the sequence is increasing.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

$$\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x$$

## Homework Equations

Maclaurin series:
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ... + (-1)^{r+1} \frac{x^r}{r} + ...$$

## The Attempt at a Solution

We're considering vanishingly small $x$, so just taking the first term in the Maclaurin series the limit becomes:

$$\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x = \lim\limits_{x \to 0} x^x = \mathrm{undefined}$$

or so I thought until google tells me that $0^0 = 1$.

What's going on here? How can I evaluate the limit properly?

Let ##F(x) = \left( \ln (1+x) \right)^x##. Then ##L(x) \equiv \ln F(x)## is given by
$$\begin{array}{rcl} L(x) &=& x \ln \left( \ln (1+x) \right) = x \ln( x - x^2/2 + x^3/3 - \cdots) \\ &=& x \ln x + x \ln ( 1 - x/2 + x^2/3 - \cdots) . \end{array}$$
Thus, ##\lim_{x \to 0} L(x) = \lim_{x \to 0} x \ln x ##. You can write
$$\lim_{x \to 0} x \ln x = \lim_{x \to 0} \frac{\ln x}{1/x},$$
and evaluate that last form using l'Hospital's rule.

member 587159
Let ##F(x) = \left( \ln (1+x) \right)^x##. Then ##L(x) \equiv \ln F(x)## is given by
$$\begin{array}{rcl} L(x) &=& x \ln \left( \ln (1+x) \right) = x \ln( x - x^2/2 + x^3/3 - \cdots) \\ &=& x \ln x + x \ln ( 1 - x/2 + x^2/3 - \cdots) . \end{array}$$
Thus, ##\lim_{x \to 0} L(x) = \lim_{x \to 0} x \ln x ##. You can write
$$\lim_{x \to 0} x \ln x = \lim_{x \to 0} \frac{\ln x}{1/x},$$
and evaluate that last form using l'Hospital's rule.

Am I missing something because I think you can't apply l'Hospital's rule because lim x>0 lnx is not defined? (We can't approach lnx from the left as it is not defined in the context of real numbers).

Ray Vickson
Homework Helper
Dearly Missed
Am I missing something because I think you can't apply l'Hospital's rule because lim x>0 lnx is not defined? (We can't approach lnx from the left as it is not defined in the context of real numbers).

The original limit cannot be taken from the left, either. When ##x < 0##, ##\ln(1+x) < 0## and ##(\ln(1+x))^x## is undefined for fractional negative ##x## (being a negative raised to a negative fractional power).

member 587159
The original limit cannot be taken from the left, either. When ##x < 0##, ##\ln(1+x) < 0## and ##(\ln(1+x))^x## is undefined for fractional negative ##x## (being a negative raised to a negative fractional power).

Well you could say that the limit does not exist then, but the OP might not have mentioned that they meant the right side limit.