# 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!

• The reciprocal of $\log(1/x)$ is $(\log(1/x))^{-1}$, not $\log((1/x)^{-1})$ as you’ve claimed. – symplectomorphic Nov 16 '19 at 17:49
• @symplectomorphic thats a great point – jeffery_the_wind Nov 17 '19 at 0:33

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:

• $$\lim_{x \to 0^+} x^x=1$$
• $$\lim_{x \to 0^+} x\log x=0$$
• $$\lim_{x \to \infty} \frac x{e^x}=0$$

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 $$x\ln x \stackrel{x\to 0^+}{\rightarrow}0$$:

$$\tan x\ln \frac{1}{x} = -\frac{\sin x}{x}\frac{1}{\cos x}\cdot x\ln x \stackrel{x \to 0^+}{\rightarrow} = -1\cdot 1\cdot 0$$ $$\Rightarrow \lim_{x \to 0^+} \left (\frac{1}{x} \right)^{\tan x}= e^0 = 1$$

• Urk, I'm sorry. You're right that that first rule is the mistake he made and he sold me on it. – Matthew Daly Nov 16 '19 at 19:02
• @trancelocation I'm just looking at your derivation there, I see 3 negatives there there but then the expression in the end is positive. Is this because $(-\log x)^{-2}=(\log x)^{-2}$? – jeffery_the_wind Nov 17 '19 at 0:42
• @jeffery_the_wind : Exactly. This is the reason. Squaring removes the sign. – trancelocation Nov 17 '19 at 6:42
• So it seems in the alternate calculation you are evaluation $- \sin x/ x$ as 1 but it takes the form 0/0 where we should use Hopitals rule but perhaps you just left that step out. – jeffery_the_wind Nov 17 '19 at 7:24
• @jeffery_the_wind : $\lim_{x\to 0}\frac{\sin x}{x}=1$ is a so called standard limit. But of course you may use L'Hospital to show it. But similarly you may need to show also that $\lim_{x\to 0^+}x\log^2x = 0$, if you do not consider this as a standard limit. – trancelocation Nov 17 '19 at 7:52

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

$$\lim_{t\to\infty}\frac{-\log t}t=-\lim_{t\to\infty}\frac1t=0.$$