Calculate $\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$ without L'Hôpital's rule

How can I calculate limit $$\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$$ without using L'Hôpital's rule?

What I have tried so far:

I tried to use the fact that $$\lim_{\alpha\to 0}(1 + \alpha)^{1/\alpha} = e$$ and do the following: $$\lim_{x\to \pi/4}\cot(x)^{\cot(4 \cdot x)} = \lim_{x\to \pi/4}(1 + (\cot(x) - 1))^{\cot(4 \cdot x)} = \lim_{x\to \pi/4}(1 + (\cot(x) - 1))^{\frac{1} {\cot(x) - 1} \cdot (\cot(x) - 1) \cdot \cot(4 \cdot x)} = \lim_{x\to \pi/4}e^{(\cot(x) - 1) \cdot \cot(4 \cdot x)} = e^{\lim_{x\to \pi/4}{(\cot(x) - 1) \cdot \cot(4 \cdot x)}}$$ But I have problems calculating limit $$\lim_{x\to \pi/4}{(\cot(x) - 1) \cdot \cot(4 \cdot x)}$$ I tried to turn $$\cot(x)$$ into $$\frac{\cos(x)}{\sin(x)}$$ as well as turning it into $$\tan(x)$$, but I do not see any workaround afterwards.

I would appreciate any pieces of advice. Thank you!

• How about a change of variable? It may clarify. – dfnu Oct 15 '19 at 11:27
• I tried to replace t = x - pi/4, thus having x = t + pi/4 and t -> 0, but unfortunately it did not clarify. Anyway, I will try to do the same thing again in case I have messed up. Thank you for your reply – alexey_zotov Oct 15 '19 at 11:33
• @alexey_zotov I think it's the right approach. Then use periodicity of cotangent and sum of angles. – dfnu Oct 15 '19 at 11:35
• Try $u = \cot(x)$; use the double-angle formula for cotangent twice to express $\cot(4x)$ using $u$. (Or use the double-angle formula for tangent, but you still want it using $u$.) Strip out the stuff that's not causing problems, and you're left with $\lim_{u\to1}u^{1/(1-u)}$. I don't know whether you're allowed to use the known value of that limit, however. – Toby Bartels Oct 15 '19 at 11:41
• By graphing, the correct answer is actually $\frac{1}{\sqrt e}$. – Toby Mak Oct 15 '19 at 12:05

$$\lim_{x\to\pi/4}\cot x^{\cot4x}=\left(\lim_{x\to\pi/4}(1+\cot x-1)^{1/(\cot x-1)}\right)^{\lim_{x\to\pi/4}{\cot4x(\cot x-1)}}$$

The inner limit converges to $$e$$

For the exponent, $$\lim_{x\to\pi/4}\cot4x(\cot x-1)=\lim_{x\to\pi/4}\dfrac{\cos4x}{\sin x}\cdot\lim_{x\to\pi/4}\dfrac{\cos x-\sin x}{\sin4x}$$

Now

$$\lim_{x\to\pi/4}\dfrac{\cos4x}{\sin x}=\dfrac{\cos\pi}{\sin\dfrac\pi4}=?$$

Finally

Method$$\#:1$$

$$F=\lim_{x\to\pi/4}\dfrac{\cos x-\sin x}{\sin4x}=\lim_{x\to\pi/4}\cos x\cdot\lim_{x\to\pi/4}\dfrac{\cot x- 1}{\sin4x}=\dfrac1{\sqrt2}\cdot\lim_{x\to\pi/4}\dfrac{\cot x- \cot\dfrac\pi4}{\sin4x-\sin\pi}$$

Method$$\#:1A$$

$$F=\dfrac1{\sqrt2}\cdot\dfrac{\dfrac{d(\cot x)}{dx}}{\dfrac{d(\sin4x)}{dx}}_{\text{at } x=\pi/4}$$

Method$$\#:1B$$

$$F=\dfrac1{\sqrt2}\cdot\lim_{x\to\pi/4}\dfrac1{\sin x\sin\dfrac\pi4} \cdot\lim_{x\to\pi/4}\dfrac{\sin\left(\dfrac\pi4-x\right)}{\sin4\left(\dfrac\pi4-x\right)}=\dfrac{\sqrt2}4$$

Method$$\#:2$$

set $$\dfrac\pi4-x=y$$ $$F=\sqrt2\lim_{x\to\pi/4}\dfrac{\sin\left(\dfrac\pi4-x\right)}{\sin4x}=\sqrt2\lim_{y\to0}\dfrac{\sin y}{\sin4\left(\dfrac\pi4-y\right)}=\dfrac{\sqrt2}4$$

• @dfnu, Please pinpoint the mistake – lab bhattacharjee Oct 15 '19 at 11:56
• You may not have $\lim f^g = (\lim f )^{\lim g}$, in particular when $\lim f = \lim g =0$, which is the case of the limits of your answer. – mathcounterexamples.net Oct 15 '19 at 11:57
• @labbhattacharjee sorry I just lost track of partial solutions. – dfnu Oct 15 '19 at 11:59
• Thank you very much! Your solution helped me a lot – alexey_zotov Oct 15 '19 at 12:13
• @mathcounterexamples.net, Here $\lim f=e\ne0$ – lab bhattacharjee Oct 15 '19 at 12:28

I would use the ansatz $$\lim_{x\to \frac{\pi}{4}}\exp(\cot(4x)\log(\cot(x)))$$ $$=\exp\left(\lim_{x\to \frac{\pi}{4}}\frac{\cos(4x)\log(\cot(x))}{\sin(4x))}\right)$$ $$=\exp\left(\lim_{x\to \frac{\pi}{4}}\cos(4x)\lim_{x\to \frac{\pi}{4}}\frac{\log(\cot(x))}{\sin(4x)}\right)$$ $$=\exp\left(-1\lim_{x\to \frac{\pi}{4}}\frac{\log(\cot(4x))}{\sin(4x)}\right)$$

We have that

$$\cot(x)^{\cot(4*x)}=\left[\left(1+(\cot(x)-1)\right)^{\frac1{\cot(x)-1}}\right]^{\cot (4x)(\cot(x)-1)} \to$$

indeed

$$\left(1+(\cot(x)-1)\right)^{\frac1{\cot(x)-1}}\to e^{-\frac12}$$

and by trigonometric identities

$$\cot (4x)(\cot(x)-1)=\frac{\cos^2 (2x)-\sin^2(2x)}{2\cos(2x)\sin (2x)}\left(\frac{1+\cos (2x)}{\sin (2x)}-1\right) \\=\frac{1-2\sin^2(2x)}{2\sin^2(2x)}\left(\frac{1+\cos (2x)-\sin(2x)}{\cos (2x)}\right)\to -\frac12$$

since by $$y=\frac{\pi}2-2x \to 0$$

$$\frac{1+\cos (2x)-\sin(2x)}{\cos (2x)}=\frac{1+\sin y-\cos y}{\sin y}=1+\frac{1-\cos y}{y^2}\frac{y}{\sin y}y\to 1+0=1$$

We substitute $$c = \cot(x)$$ and we get $$\lim\limits_{x\to\pi/4} \cot(x)^{\cot(4x)} =\lim\limits_{c\to 1}\; c^{\frac{c^4-6c^2+1}{4c^3-4c}}$$ because $$\cot(4x) = \frac{\cot^4x-6\cot^2x+1}{4\cot^3x-4\cot x}$$ Then we have $$\lim\limits_{c\to 1} \;c^{\frac{c^4-6c^2+1}{4c^3-4c}} =\lim\limits_{c\to 1} \;\exp\left(\ln(c) \cdot \frac{c^4-6c^2+1}{4c^3-4c}\right) =\lim\limits_{c\to 1} \;\exp\left(\frac{\ln(c)}{c-1} \cdot \frac{c^4-6c^2+1}{4c^2+4c}\right) \\ =\exp\left(\lim\limits_{c\to 1} \;\frac{\ln(c)}{c-1} \cdot \lim\limits_{c\to 1} \;\frac{c^4-6c^2+1}{4c^2+4c}\right) =\exp\left(1\cdot\frac{-4}{8} \right)= \frac{1}{\sqrt{e}}$$ because $$\lim\limits_{c\to 1} \;\frac{\ln(c)}{c-1}$$ is the derivative of $$\ln(x)$$ evaluated at $$x=1$$, which is $$1$$.

• It's an interesting and profound solution, thank you! – alexey_zotov Oct 15 '19 at 12:15
• @alexey_zotov Did you notice that my result is different from the one you accepted? – Reinhard Meier Oct 15 '19 at 12:17
• The accepted answer did not provide the result, only the in-between numbers. By multiplying them, I got the same result. I accepted it because it developed the ideas I had already been trying to realize – alexey_zotov Oct 15 '19 at 12:21
• @alexey_zotov Oops, I missed that. It is not obvious at the first glance how the results are related. – Reinhard Meier Oct 15 '19 at 12:28

With a subtitution $$u = x-\frac{\pi}4$$ you get

$$\begin{eqnarray}\lim_{u\to 0} \left[\frac{\cos\left(u+\frac{\pi}4\right)}{\sin\left(u+\frac{\pi}4\right)} -1\right]\cot(4u+\pi).\end{eqnarray}$$

Using periodicity and sum of angles you now have

$$\lim_{u\to 0} \left[\frac{\frac{\sqrt{2}}2\cos u - \frac{\sqrt 2}{2}\sin u}{\frac{\sqrt{2}}2\cos u + \frac{\sqrt 2}{2}\sin u} -1\right]\cot(4u).$$

Least common denominator brings you to

$$\lim_{u\to 0} \frac{-\sqrt 2 \sin u}{\frac{\sqrt{2}}2\cos u + \frac{\sqrt 2}{2}\sin u}\cdot \frac{\cos 4u}{\sin 4u}=\frac{-\sqrt 2}{\frac{\sqrt 2}2}\cdot \frac14=-\frac12.$$

Thus your limit is $$\frac1{\sqrt e}.$$

• Understand! I'll delete the comment. – mathcounterexamples.net Oct 15 '19 at 12:10

You have $$\cot(x+y) = \frac{\cot x \cot y -1}{\cot x + \cot y}$$

Taking $$h = x-\pi/4$$ you get

$$\cot x = \cot(\pi/4 + h) = \frac{\cot h -1}{\cot h + 1} = \frac{\cos h - \sin h}{\cos h + \sin h} = 1- 2h +o(h)$$ around zero.

And with $$4x = \pi + 4h$$

$$\cot(4x) = \cot(\pi + 4h) = \frac{\cos(\pi+4h)}{\sin(\pi+4h)} = \frac{\cos(4h)}{\sin(4h)} = \frac{1}{4h} +o\left(\frac{1}{h}\right)$$

Now $$\lim\limits_{h \to 0} (1-2h)^{\frac{1}{4h}} = 1/\sqrt{e}$$ which is the limit you're looking for.

• @ReinhardMeier Thanks for spotting the mistake. I'll update the answer. – mathcounterexamples.net Oct 15 '19 at 12:07
• I have deleted my comment now that you have updated the answer. – Reinhard Meier Oct 15 '19 at 12:29