L'Hôpital's rule exercise with natural log function I'm looking for some advice on the following exercise:
$$\lim_{x \to 0^+}{\ln{(\frac{1}{x}})}^x$$
This is my work so far:
$$\lim_{x \to 0^+}{\ln{(\frac{1}{x}})}^x = \lim_{x \to 0^+}{x\cdot\ln{(\frac{1}{x}})} = \lim_{x \to 0^+}{\frac{\ln{\frac{1}{x}}}{\frac{1}{x}}}$$
Taking the derivatives of the top and bottom:
$$\frac{d}{dx}\ln{\frac{1}{x}}=-\frac{1}{x}\,\,\text{and}\,\,\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$$
So:
$$\lim_{x \to 0^+}{\frac{-\frac{1}{x}}{-\frac{1}{x^2}}} = \lim_{x \to 0^+}{\frac{x^2}{x}}=\lim_{x \to 0^+}{x} = 0$$
...
Which doesn't quite add up.  I'm thinking, I should rewrite the equation in the form of $e^x$ as the limit would then approach $1$, which is the correct answer.  However, I'm a bit stuck on that front.
Any suggestions would be greatly appreciated!
 A: $$\lim_{x\to 0}\ln\left(\frac{1}{x}\right)^x\neq \lim_{x\to 0}x\ln\left(\frac{1}{x}\right)$$
You have that $$\ln\left(\frac{1}{x}\right)^x=e^{x\ln(\frac{1}{x})}.$$
Since $$\lim_{x\to 0}x\ln\left(\frac{1}{x}\right)=\lim_{x\to 0}\frac{\ln\left(\frac{1}{x}\right)}{\frac{1}{x}}\underset{\text{(Hop.)}}{=}\lim_{x\to 0}\frac{\frac{\frac{-1}{x^1}}{\frac{1}{x}}}{-\frac{1}{x^2}}=\lim_{x\to 0} x=0$$
and that $x\mapsto e^x$ is continuous at $x=0$, we can conclude that $$\lim_{x\to 0}\ln\left(\frac{1}{x}\right)^x=e^0=1.$$
A: One should be careful about the exponent, there are two possible meanings:
\begin{equation*}
\ln \left( \left( \frac{1}{x}\right) ^{x}\right) \neq \left( \ln \left( 
\frac{1}{x}\right) \right) ^{x}
\end{equation*}
Let us compute the limit of both of them.
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\ln \left( \left( \frac{1}{x}\right) ^{x}\right)
=\lim_{x\rightarrow 0^{+}}x\ln \left( \frac{1}{x}\right) =\lim_{x\rightarrow
0^{+}}x\left( \ln 1-\ln x\right) =-\lim_{x\rightarrow 0^{+}}x\ln x=0.
\end{equation*}
However,
\begin{eqnarray*}
\lim_{x\rightarrow 0^{+}}\left( \ln \left( \frac{1}{x}\right) \right) ^{x}
&=&\lim_{x\rightarrow 0^{+}}e^{\ln \left( \ln \left( \frac{1}{x}\right)
\right) ^{x}} \\
&=&\lim_{x\rightarrow 0^{+}}e^{x\ln \left( \ln \left( \frac{1}{x}\right)
\right) } \\
&=&\lim_{x\rightarrow 0^{+}}e^{\frac{\ln \left( \ln \left( \frac{1}{x}%
\right) \right) }{\left( \frac{1}{x}\right) }} \\
&=&\lim_{x\rightarrow 0^{+}}e^{\frac{\left[ \ln \left( \ln \left( \frac{1}{x}%
\right) \right) \right] ^{\prime }}{\left( \frac{1}{x}\right) ^{\prime }}} \\
&=&\lim_{x\rightarrow 0^{+}}e^{\frac{\frac{\left( \ln \left( \frac{1}{x}%
\right) \right) ^{\prime }}{\left( \ln \left( \frac{1}{x}\right) \right) }}{-%
\frac{1}{x^{2}}}} \\
&=&e^{-\lim\limits_{x\rightarrow 0^{+}}\frac{\left( -\frac{1}{x}\right) }{%
\ln \left( \frac{1}{x}\right) }x^{2}} \\
&=&e^{-\lim\limits_{x\rightarrow 0^{+}}\frac{x}{\ln x}} \\
&=&e^{0}=1.
\end{eqnarray*}
A: Like many have pointed out the confusion is because of the notation $\log a^{b}$. The accepted convention in most textbooks is that $\log a^{b} = \log (a^{b})$ and not $(\log a)^{b}$. However to avoid any confusion it is easy to handle both the interpretations of $\log a^{b}$.
Thus we can calculate the limit of $f(x) = \log((1/x)^{x})$ as follows. Clearly in this case we have $f(x) = x\log (1/x) = -x\log x$ and putting $x = 1/y$ we can see that we have $f(x) = (\log y)/y$ and $y \to \infty$. This is a standard limit whose value is $0$, but it worthwhile to show the calculation. Since $y \to \infty$ we can assume that $y > 1$ and hence $\sqrt{y} > 1$. We have the standard inequality $$0 \leq \log t \leq t - 1$$ for $t \geq 1$ and putting $t = \sqrt{y}$ we get $$0 \leq \log\sqrt{y} \leq \sqrt{y} - 1 < \sqrt{y}$$ or $$0 \leq \frac{1}{2}\log y < \sqrt{y}$$ or $0 \leq \log y < 2\sqrt{y}$ and hence $0 \leq (\log y)/y < 2/\sqrt{y}$ Now letting $y \to \infty$ and using Squeeze theorem we get $\lim_{y \to \infty}(\log y)/y = 0$.
Again if we consider $\log(1/x)^{x}$ as $(\log(1/x))^{x}$ then we have $$f(x) = \exp(x\log(\log(1/x)))$$. Putting $y = 1/x$ so that $y \to \infty$ as $x \to 0^{+}$ we can see that $$f(x) = \exp((\log\log y)/y) = \exp(g(y))$$ Clearly we have $$g(y) = \frac{\log \log y}{y} = \frac{\log \log y}{\log y}\cdot\frac{\log y}{y} \to 0\cdot 0 = 0$$ as $y \to \infty$ (from previous paragraph). And hence $f(x) = \exp(g(y)) \to \exp(0) = 1$ as $y \to \infty$.
To summarize
\begin{align}
\lim_{x \to 0^{+}}\log\left(\left(\frac{1}{x}\right)^{x}\right) &= 0\notag\\
\lim_{x \to 0^{+}}\left(\log\left(\frac{1}{x}\right)\right)^{x} &= 1
\end{align}
Note that when $x < 0$ these functions are undefined at infinitely many points in any neighborhood of $0$ and hence their limits as $x \to 0^{-}$ can't be considered.
