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$$\lim\limits_{x\to\frac{\pi}{2}} \frac{a^{\cos x} - 1}{ \cos x}$$

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closed as off-topic by Nosrati, Leucippus, Theoretical Economist, Holo, Arnaud D. Sep 26 '18 at 22:03

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As $\cos (\pi / 2) = 0$, then your limit is equivalent to

$$ \lim_{x \to \pi/2} \frac{ a^{\cos x } - a^{\cos \pi/2}}{\cos x - \cos (\pi/2) }\cdot \frac{x-\pi/2}{x-\pi/2} = \frac{ (a^{\cos x })'}{(\cos x)'} \bigg|_{x = \pi/2} $$

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We are assuming $a>0$, let $y=\cos x\to 0$ and use the definition of derivative, that is

$$\lim\limits_{x\to\frac{\pi}{2}} \frac{a^{\cos x} - 1}{ \cos x}=\lim\limits_{y\to0} \frac{a^{y} - 1}{ y}$$

then recall that by $f(y)=a^y \implies f'(y)=a^y \ln a\,$ we have

$$\lim\limits_{y\to0} \frac{a^{y} - 1}{ y}=\lim\limits_{y\to0} \frac{f(y) - f(0)}{ y-0}=f'(0)=\ln a$$

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By L'Hospital's Rule, $$\displaystyle \lim_{x\rightarrow\frac{\pi}{2}} \frac{a^{\cos(x)}-1}{\cos(x)}=\lim_{x\rightarrow\frac{\pi}{2}} \frac{(a^{\cos x}-1)^{'}}{(\cos x)^{'}},$$ which is equal to $$\lim_{x\rightarrow\frac{\pi}{2}} \frac{-(\sin x)\ln(a)a^{\cos x}}{-\sin x} = \ln(a)$$

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