# How do I find this limit?

I am a little bit confused. How can I find the following limit?

$$\lim_{x\to \pi/2} (\tan x)^{\tan 2x}$$ Seems like infinity to the power of $0$.

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You probably want $x\to\frac{\pi-}{2}$ (left hand limit) rather than $x\to\pi/2$? – user1551 Jan 5 '13 at 10:01
@BabakSorouh the 2 answers below are actually what I meant, seems like working – Ada Jan 5 '13 at 10:06
@Ada: Yes! I made a mistake in calculating something there. Thanks. – Babak S. Jan 5 '13 at 10:08

As the base needs to be positive for such exmponentation to be well defined, you need to evaluate $$\lim_{x\to \pi/2^-} (\tan x)^{\tan 2x}=\lim_{x\to \pi/2^-} e^{\ln(\tan x)\tan 2x}$$ This reduces to $$\lim_{x\to \pi/2^-} \ln(\tan x)\tan 2x=\lim_{x\to \pi/2^-} [\ln(\sin x)-\ln(\cos x)]\tan 2x=0-\lim_{x\to \pi/2^-} \ln(\cos x)\tan 2x$$ So, $$-\lim_{x\to \pi/2^-} \ln(\cos x)\tan 2x=-\lim_{x\to \pi/2^-} \ln(\cos x)\frac{\sin 2x}{\cos 2x}=\lim_{x\to \pi/2^-} \ln(\cos x)\sin 2x=\\\lim_{x\to \pi/2^-} 2\sin x\ln(\cos x)\cos x =2\lim_{x\to \pi/2^-} \ln(\cos x)\cos x=2\lim_{x\to \pi/2^-} \frac{\ln(\cos x)}{\frac1{\cos x}}$$ The last limit shouldn't be hard to compute with De L'Hopital

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It is essential for the OP that $x\to\pi^-/2$. – Babak S. Jan 5 '13 at 10:10

Let

$$L = \lim_{x \rightarrow \pi/2} (\tan{x})^{\tan{2 x}}$$

$$\log{L} = \lim_{x \rightarrow \pi/2} \tan{2 x} \log{(\tan{x})}$$

$$= \lim_{x \rightarrow \pi/2} \frac{\log{(\tan{x})}}{\cot{2 x}}$$

Now use L'Hopital's Rule, as the limit is of type $\infty/\infty$:

$$\log{L} = -\lim_{x \rightarrow \pi/2} \frac{1}{\tan{x}} \frac{\sec^2{x}}{2 \csc^2{2 x}}$$

$$= -\lim_{x \rightarrow \pi/2} \frac{\cos{x}}{\sin{x}} \frac{2 \sin^2{x} \cos^2{x}}{\cos^2{x}}$$

$$= -\lim_{x \rightarrow \pi/2} 2 \cos{x} \sin{x} = 0$$

Therefore, $L = 1$.

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Hint: Besides to @Nameless's answer. Set $\pi/2-x=t$ and use this fact that $\tan(\pi/2-x)=\cot(x), \tan(\pi+x)=\tan(x)$ and when $x$ tends to zero we khnow that $\tan(x)\sim x$. I am sure this approach also works.

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Please look at my answer that differ from others. Can you see, is there any mistake – Adi Dani Jan 5 '13 at 11:58
Nice approach! +1 – amWhy Feb 23 '13 at 14:03
@amWhy: Thanks Amy. mornin. :) – Babak S. Feb 23 '13 at 14:32
Good morning (or should I say, good afternoon! to you?). Nice to "see" your "smiling face". BTW: I love Peter Cameron too!! – amWhy Feb 23 '13 at 14:53
@amWhy: Or say, have a nice dream! $\ddot\smile$ – Babak S. Feb 23 '13 at 16:19

Let $t=\tan x$. Using the double angle formula for tangent, we get $$\lim_{x\to{\pi/2}^-} (\tan x)^{\tan 2x} =\lim_{t\to+\infty}t^{2t/(1-t^2)} =\lim_{t\to+\infty}\exp\left(\frac{2t\log t}{1-t^2}\right) =\exp\left(\lim_{t\to+\infty}\frac{2t\log t}{1-t^2}\right).$$ Using L'Hospital rule, we have $$\lim_{t\to+\infty}\frac{2t\log t}{1-t^2} = \lim_{t\to+\infty}\frac{1+\log t}{-t} = 0.$$ Hence the left hand limit is $1$.

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I like this one.+ – Babak S. Jan 5 '13 at 15:57
(+1) nice technique. – Mhenni Benghorbal Jan 5 '13 at 16:27

let y=〖(tan〗⁡〖x)〗^tan2x ln⁡〖y=〗 tan⁡2x.ln⁡(tanx) =(sin⁡2 x.ln⁡(tanx))/cos2x lim┬(x→π/4)⁡lny=lim┬(x→π/4) ln⁡(tanx)/cot2x =0/0 by lop ln⁡〖lim┬(x→π/4) y〗=lim┬(x→π/4) ⁡〖sec^2⁡x/(-2 csc^2⁡2x tanx) 〗=-1 lim┬(x→π/4) y=1/e

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