Evaluating $\lim_{x \to 0^+} \left (\frac{1}{x} \right)^{\tan x}$ with Hôpital's rule Again, hoping someone can help me check my work here. I need to evaluate this limit:
$$\lim_{x \to 0^+} \left (\frac{1}{x} \right)^{\tan x}$$
So we take the natural logarithm
$$\lim_{x \to 0^+} \log \left (\frac{1}{x} \right)^{\tan x}=\lim_{x \to 0^+} \tan x \log \left (\frac{1}{x} \right) =\lim_{x \to 0^+} \frac{\tan x}{\frac{1}{\log \left (\frac{1}{x} \right)}}$$
The limits of numerator and denominator are zero, so we can apply Hôpitals Rule. Notice that
$$\frac{d}{dx}\tan x =\sec^2x$$
and
$$\frac{d}{dx}\frac{1}{\log\left(\frac{1}{x}\right)}= \frac{d}{dx} \log\left(\frac{1}{x}\right)^{-1}= \frac{d}{dx} -(\log(1)-\log(x))= \frac{d}{dx} \log(x)=\frac{1}{x}$$
* Update *
I had made a mistake on this previous step, here is the correction, as pointed out in the comments.
$$\frac{d}{dx}\frac{1}{\log\left(\frac{1}{x}\right)}= \frac{d}{dx} \left(\log\frac{1}{x}\right)^{-1}= \frac{d}{dx} (\log(1)-\log(x))^{-1}= \frac{d}{dx} -\log(x)^{-1}=-1(- \log x )^{-2} \frac{-1}{x}=\frac{1}{x \log^2 x}$$
so now
$$\lim_{x \to 0^+} \log \left (\frac{1}{x} \right)^{\tan x}=\lim_{x \to 0^+} \frac{\tan x}{\frac{1}{\log \left (\frac{1}{x} \right)}}=\lim_{x \to 0^+} \frac{\sec^2 x}{\frac{1}{x \log^2 x}}=\lim_{x \to 0^+} x \log^2 x \sec^2x=0$$
$$\implies \lim_{x \to 0^+} \left (\frac{1}{x} \right)^{\tan x}=\lim_{x \to 0^+}e^{ \log \left (\frac{1}{x} \right)^{\tan x}}=e^{0}=1$$
Thanks!
 A: Edit: After your correction concerning


*

*$\left(\frac{1}{\log \frac 1x} \right)' = \frac{1}{x\log^2x}$
your calculation is correct now. 
As limits can often be calculated in several ways "appealing to different tastes" - here is another way using $x\ln x \stackrel{x\to 0^+}{\rightarrow}0$:
$$\tan x\ln \frac{1}{x} = -\frac{\sin x}{x}\frac{1}{\cos x}\cdot x\ln x \stackrel{x \to 0^+}{\rightarrow} = -1\cdot 1\cdot 0$$
$$\Rightarrow \lim_{x \to 0^+} \left (\frac{1}{x} \right)^{\tan x}= e^0 = 1$$
A: Your work is correct, but you can get rid of the tangent by writing
$$\tan x=x\frac{\tan x}x$$ and the fraction is known to tend to $1$.
Hence
$$\lim_{x\to0^+}\left(\frac1x\right)^{\tan x}=\left(\lim_{x\to0^+}\left(\frac1x\right)^x\right)^{\lim_{x\to0^+}\tan x/x}=\frac1{\lim_{x\to0^+}x^x}.$$
Then $\log x^x=x\log x$ can be processed by L'Hospital as you did, or as
$$\lim_{t\to\infty}\frac{-\log t}t=-\lim_{t\to\infty}\frac1t=0.$$
A: Since
$$\left (\frac{1}{x} \right)^{\tan x}=\frac{1^{\tan x}}{x^{\tan x}} =\frac{1}{x^{\tan x}}$$
we need to evaluate
$$\lim_{x \to 0^+} x^{\tan x}=\lim_{x \to 0^+} e^{(\tan x\cdot \log x)}=1$$
indeed by standard limits
$$\tan x\cdot \log x=\frac{\tan x}x \cdot x\log x \to 1 \cdot 0 =0$$
indeed by $x=e^{-y}\to 0^+$ with $y\to \infty$
$$ x\log x=e^{-y}\log (e^{-y})=-\frac y{e^y}\to 0$$
which can be easily proved bu l'Hospital or observing that eventually $e^y\ge y^2$ and thus
$$\frac y{e^y}\le \frac y{y^2}=\frac 1y\to 0$$
Finally try to keep in mind for the future the foundamental equivalent results here obtained:


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*$\lim_{x \to 0^+} x^x=1$

*$\lim_{x \to 0^+} x\log x=0$

*$\lim_{x \to \infty} \frac x{e^x}=0$
