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$$ \lim_{x \rightarrow 0} [\tan(x) + \sec(x)]^{\csc(x)} = e $$

how to arrive at e, according to wolfram alpha, that this is the answer?

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    $\begingroup$ The values of $\tan$ and $\sec$ at $0$ are respectively $0$ and $1,$ and the slopes are respectively $1$ and $0,$ so $\tan x + \sec x$ is like $x+1,$ and the value of $\sin$ at $0$ is $0$ and its slope at $0$ is $1,$ so $1/\sin x$ is like $1/x.$ Hence it is not surprising that $$ \lim_{x\to0} (\tan x + \sec x)^{1/\sin x} = \lim_{x\to0} (x+1)^{1/x}. $$ $\endgroup$ – Michael Hardy Jun 5 '18 at 2:35
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    $\begingroup$ In fact $$\tan x + \sec x = 1 + x + \text{higher-degree terms} $$ and $$\frac 1 {\sin x} = \left( \frac 1 x + \text{a power series in } x\right). $$ $\endgroup$ – Michael Hardy Jun 5 '18 at 2:37
  • $\begingroup$ BTW one of the tangent half-angle formulas is $$ \tan x + \sec x = \tan\left( \frac x 2 + \frac \pi 4\right). $$ Privately I think of this as the "cartographer's tangent half-angle formula". I don't know if anyone else uses that term. $\endgroup$ – Michael Hardy Jun 5 '18 at 2:39
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As $ x \to 0$, $\tan(x) + \sec(x) = 1 + x + O(x^2)$ while $\csc(x) = 1/x + O(x)$. Thus $$\ln\left((\tan(x)+\sec(x))^{\csc(x)}\right) = \csc(x) \ln(\tan(x)+\sec(x)) = \left(\frac{1}{x} + O(x)\right)(x + O(x^2)) = 1 + O(x)$$ and $$(\tan(x)+\sec(x))^{\csc(x)} = \exp(1+O(x)) = e + O(x)$$

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    $\begingroup$ Wow!! I’ve just started to grasp asymptotic analysis and big-$O$ series expansions . . . this is marvelously ingenuitive and elegant! $\endgroup$ – gen-z ready to perish Jun 5 '18 at 3:07
  • $\begingroup$ I’m curious, actually: why can you use a big-$O$ expansion when, I thought, those were for $x\to\infty$? $\endgroup$ – gen-z ready to perish Jun 5 '18 at 4:16
  • $\begingroup$ the Taylor series is by definition infinite $\endgroup$ – user29418 Jun 5 '18 at 4:18
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    $\begingroup$ You can have $O()$ associated to any kind of limits. These are for $x \to 0$. $\endgroup$ – Robert Israel Jun 5 '18 at 5:33
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Let $x=\arcsin(t)$. Then, we have

$$\begin{align} \left(\tan(x)+\sec(x)\right)^{1/\sin(x)}&=\left(\frac{1+t}{\sqrt{1-t^2}}\right)^{1/t}\\\\ &=\left(\frac{1+t}{1-t}\right)^{1/(2t)}\\\\ \end{align}$$


Now, letting $u=1/t$ reveals

$$\begin{align} \left(\tan(x)+\sec(x)\right)^{1/\sin(x)}&=\left(\frac{u+1}{u-1}\right)^{u/2}\\\\ &=\left(1+\frac{2}{u-1}\right)^{u/2} \end{align}$$


Finally, enforcing the substitution $v=(u-1)/2$ yields

$$\left(\tan(x)+\sec(x)\right)^{1/\sin(x)}=\left(1+\frac1{v}\right)^{v}\sqrt{1+\frac1{v}}$$


As $x\to 0$, $v\to \infty$ and we have

$$\lim_{x\to 0}\left(\tan(x)+\sec(x)\right)^{1/\sin(x)}=\lim_{v\to\infty}\left(1+\frac1{v}\right)^{v}\sqrt{1+\frac1{v}}=e$$

as was to be shown!

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Note that: $$\begin{align}\lim_{x \rightarrow 0} [\tan(x) + \sec(x)]^{\csc(x)} = &\lim_{x \rightarrow 0} \left[\frac{\sin x}{\cos x} + \frac{1}{\cos x}\right]^{1/\sin x} =\\ &\lim_{x \rightarrow 0} \frac{[1+\sin x]^{1/\sin x}}{[\cos x]^{1/\sin x}} =\\ &\frac{\lim_{x \rightarrow 0}[1+\sin x]^{1/\sin x}}{\lim_{x \rightarrow 0}[\cos x]^{1/\sin x}} =\\ &\frac{e}{1}=e.\end{align}$$

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$$(\sec x+\tan x)^{\csc x}=\left(\left(1+\sec x+\tan x-1\right)^{1/(\sec x+\tan x-1)}\right)^{(\sec x+\tan x-1)/\sin x}$$

Use https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution for the outer exponent

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Using the same approach as in Robert Israel's anwser, consider $$A=(\tan (x)+\sec (x))^{\csc (x)}\implies \log(A)=\csc (x)\log (\tan (x)+\sec (x))$$ Now, using Taylor series built at $x=0$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$ $$\sec(x)=1+\frac{x^2}{2}+\frac{5 x^4}{24}+O\left(x^5\right)$$ $$\tan (x)+\sec (x))=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{5 x^4}{24}+O\left(x^5\right)$$ $$\log (\tan (x)+\sec (x))=x+\frac{x^3}{6}+O\left(x^5\right)$$ $$\csc(x)=\frac{1}{x}+\frac{x}{6}+\frac{7 x^3}{360}+O\left(x^5\right)$$ $$\log(A)=1+\frac{x^2}{3}+O\left(x^4\right)$$ $$A=e^{\log(A)}=e+\frac{e x^2}{3}+O\left(x^4\right)$$ which is a quite good approximation of the original function (just for suriosity, plot both of them on the same graph for $0 \leq x \leq 0.5$).

Try fro $x=\frac \pi {6}$ for which we know the exact values of the trigonometric functions. The exact value is $A=3$ while the approximation gives $e+\frac{e \pi ^2}{108}\approx 2.9667$.

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