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!
Best Answer
Edit: After your correction concerning
your calculation is correct now.
As limits can often be calculated in several ways "appealing to different tastes" - here is another way using $x\ln x \stackrel{x\to 0^+}{\rightarrow}0$:
$$\tan x\ln \frac{1}{x} = -\frac{\sin x}{x}\frac{1}{\cos x}\cdot x\ln x \stackrel{x \to 0^+}{\rightarrow} = -1\cdot 1\cdot 0$$ $$\Rightarrow \lim_{x \to 0^+} \left (\frac{1}{x} \right)^{\tan x}= e^0 = 1$$