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Could anyone help me with this trigonometric limit? I'm trying to evaluate it using L'Hôpital's rule.

I tried the following: $y = x - \frac{\pi}{2}$ and as x approaches $\frac{\pi}{2}$, $y$ approaches $0$ so I got $\lim_{x\to \frac{\pi}{2}^-}{(\tan (y + \frac{\pi}{2}))^y}$

I am stuck at deriving to get a $\frac{0}{0}$

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  • $\begingroup$ Thank you for pointing out the type error, I've edited my original question $\endgroup$ – Peter Mar 6 '15 at 0:59
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    $\begingroup$ There is one thing which is interesting (if you enjoy complex numbers) : how is approached the limit when $x$ approaches $\frac{\pi}{2}$ ? Taylor would give $$\big(\tan (x)\big)^{x-\frac{\pi}{2}}=1+\left(x-\frac{\pi }{2}\right) \left(-\log \left(x-\frac{\pi }{2}\right)+i \pi \right)$$ $\endgroup$ – Claude Leibovici Mar 6 '15 at 7:48
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Here are the steps $$ \lim\limits_{x\to\frac{\pi}{2}^-} \left(\tan x\right)^{x-\frac{\pi}{2}} $$ Let $z=x-\frac{\pi}{2}$, so now we have $$\lim\limits_{z\to 0^-} \left(\tan\left(z+\frac{\pi}{2}\right)\right)^{z}=\lim\limits_{z\to 0^-} \left(-\cot z\right)^{z}$$ $$=\lim\limits_{z\to 0^-} \exp\left(\ln\left(-\cot z\right)^{z}\right)=\lim\limits_{z\to 0^-} \exp\left(z\ln\left(-\cot z\right)\right)$$ $$ =\exp\left(\lim\limits_{z\to 0^-} \left[z\ln\left(-\cot z\right)\right]\right)=\exp\left(\lim\limits_{z\to 0^-} \left[\frac{\ln\left(-\cot z\right)}{\dfrac{1}{z}}\right]\right) $$ $$ = \exp\left(\lim\limits_{z\to 0^-} \left[\frac{\frac{d}{dz}[\ln\left(-\cot z\right)]}{\frac{d}{dz}\left[\dfrac{1}{z}\right]}\right]\right)= \exp\left(\lim\limits_{z\to 0^-} \left[\frac{-\csc(z)\sec(z)}{-\dfrac{1}{z^2}}\right]\right) $$ $$ = \exp\left(\lim\limits_{z\to 0^-} \left[z^2\csc(z)\sec(z)\right]\right)= \exp\left(\lim\limits_{z\to 0^-} \left[\frac{z^2}{\sin(z)\cos(z)}\right]\right) $$ $$= \exp\left(\lim\limits_{z\to 0^-} \left[\frac{z^2}{\sin(z)}\right]\cdot \lim\limits_{z\to 0^-}\left[\frac{1}{\cos(z)}\right]\right)=\exp\left(\lim\limits_{z\to 0^-} \left[\frac{\frac{d}{dz}\left[z^2\right]}{\frac{d}{dz}[\sin(z)]}\right]\cdot \left[\frac{1}{\cos(0)}\right]\right) $$ $$ =\exp\left(\lim\limits_{z\to 0^-} \left[\frac{2z}{\cos(z)}\right]\cdot 1\right) =\exp\left(\frac{0}{1}\right)=\exp\left(0\right)=1 $$

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Hint: First apply $\log$ to the expression in the limit.

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make a change of variable $$\pi/2 - x = 1/u, y = (\tan x )^{x-\pi/2} = \left(\tan(\pi/2 - 1/u)\right)^{-1/u} = \left(\frac{\cos 1/u}{\sin 1/u}\right)^{-1/u}$$ taking the logarithm gives $$\ln y = -\frac{\ln(u+ \cdots )}{u} \to 0 \text{ as } u \to \infty. $$

therefore $$\lim_{x \to \pi/2-} (\tan x )^{\pi/2 - x} = 1$$

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  • $\begingroup$ Should that be $$x-π/2$$ ? $\endgroup$ – Peter Mar 6 '15 at 3:16
  • $\begingroup$ @Peter, yes. does not alter the occlusion but i have edited my post. $\endgroup$ – abel Mar 6 '15 at 3:20

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