We begin by considering the limit: $$\lim_{\displaystyle x \to \frac{\pi}{2}}{\frac{\cos{x}}{\sin{x}-1}}$$ By applying L' Hôspital's rule to each side of the limit, we can prove that they are different and so the limit therefore does not exist. Furthermore, it can be proven like this: $$\lim_{\displaystyle x \to \frac{\pi}{2}}{\frac{\cos{x}}{\sin{x}-1}}=\lim_{\displaystyle x \to \frac{\pi}{2}}{\frac{\cos{x}(\sin{x}+1)}{-\cos^2{x}}}=\lim_{\displaystyle x \to \frac{\pi}{2}}{\frac{\sin{x}+1}{-\cos{x}}}$$ and now by checking each side of the limit, I have definitively concluded that the limit does not exist. However, I got curious and tried a $u$-substitution, by setting $u=\displaystyle x-\frac{\pi}{2}$ and then of course $u_0=\lim_{x \to \frac{\pi}{2}}{\left(\displaystyle x-\frac{\pi}{2}\right)}=0$ and so the limit becomes: $$\lim_{\displaystyle x \to \frac{\pi}{2}}{\frac{\cos{x}}{\sin{x}-1}}=\lim_{u \to 0}{\frac{\sin{x}}{-\cos{x}-1}}=0$$ What is the substitution not valid in the last method or why does it not work?
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1$\begingroup$ Do you mean, $$\lim_{\displaystyle x \to \frac{\pi}{2}}{\frac{\cos{x}}{\sin{x}-1}}=\lim_{u \to 0}{\frac{\sin{u}}{-\cos{u}-1}}=0$$, and if so, where did $\lim_{u \to 0}{\frac{\sin{u}}{-\cos{u}-1}}=0$ come from? $\endgroup$– Adam RubinsonCommented Jan 29, 2022 at 15:21
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$\begingroup$ You may also be interested in math.stackexchange.com/questions/1342202/… $\endgroup$– pre-kidneyCommented Jan 29, 2022 at 15:22
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$\begingroup$ The last limit should be that for $\frac{\sin u}{\cos u-1}$ $\endgroup$– blamocurCommented Jan 29, 2022 at 15:25
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$\begingroup$ Nomenclature: L'Hôpital's rule, or L'Hospital's rule, but not L'Hôspital's rule. $\endgroup$– ComptonScatteringCommented Jan 30, 2022 at 0:11
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$\begingroup$ Not strictly relevant to your question, but please do not put \displaystyle in the bottom of \lim. That is really a pain to the eye. Write instead simply \lim_{x\to\pi/2} to get $$\lim_{x\to\pi/2}$$ which looks way better. $\endgroup$– YiFan TeyCommented Jan 30, 2022 at 12:23
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Your substitution was incorrect. $$\lim_{ x \to \frac{\pi}{2}}{\frac{\cos{x}}{\sin{x}-1}}=\lim_{u \to 0}{\frac{\cos(\color{red}{u+\frac{\pi}2})}{\sin(\color{red}{u+\frac{\pi}2})-1}}=\lim_{u \to 0}{\frac{-\sin{u}}{\cos{u}-1}}$$
which yields the expected result
