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There is a problem that is easy to solve with L'hospital but we are required to solve it without it, but I could not find the answer. x* sinx part is especially confusing me, because other examples I have solved did not include such part.

$$\lim_{x \to π/2}\frac{2x\sin(x) - π}{\cos x}$$

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Define $f(x) = 2x\sin x, g(x) = \cos x.$ The expression equals

$$\frac{f(x) - f(\pi/2)}{g(x) - g(\pi/2)} = \dfrac{ \dfrac{f(x) - f(\pi/2)}{x-\pi /2}} { \dfrac{g(x) - g(\pi/2)}{x-\pi/2}}.$$

By definition of the derivative, the last expression $\to f'(\pi/2)/g'(\pi/2),$ and an easy computation shows this equals $-2.$ (And no, we did not use L'Hopital.)

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  • $\begingroup$ You're correct; you didn't apply LHR. Yet, the mechanics are effectively the same. ;-) $\endgroup$ – Mark Viola Nov 11 '17 at 21:33
  • $\begingroup$ Reciting the derivation of the special case of l'Hopital's rule is using l'Hopital. $\endgroup$ – Eric Towers Nov 11 '17 at 21:35
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    $\begingroup$ True, but this method can be used very soon in a calculus course, while L'Hopital has to wait. I never get tired of pointing this out, do I? $\endgroup$ – zhw. Nov 11 '17 at 21:35
  • $\begingroup$ @zhw. : Perhaps you should. $\endgroup$ – Eric Towers Nov 11 '17 at 21:41
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Let $t=x-\pi/2$. Then, we have

$$\begin{align} \lim_{x\to \pi/2}\frac{2x\sin(x)-\pi}{\cos(x)}&=\lim_{t\to 0}\frac{2(t+\pi/2)\sin(t+\pi/2)-\pi}{\cos(t+\pi/2)}\\\\ &=\lim_{t\to 0}\frac{\pi(1-\cos(t))-2t\cos(t)}{\sin(t)}\\\\ &=\lim_{t\to 0}\left(2\pi \frac{\sin^2(t/2)}{\sin(t)}-2\,\frac{\cos(t)}{\sin(t)/t}\right)\\\\ &=\lim_{t\to 0}\left(\pi \frac{\sin(t/2)}{\cos(t/2)}-2\,\frac{\cos(t)}{\sin(t)/t}\right)\\\\ &=-2 \end{align}$$

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  • $\begingroup$ ... I did the same thing ... $\endgroup$ – user8277998 Nov 11 '17 at 21:24
  • $\begingroup$ @123 That is not evident. $\endgroup$ – Mark Viola Nov 11 '17 at 21:31
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$$\begin{align}\lim_{x \to π/2}\frac{2x\sin(x) - π}{\cos x} = &\lim_{x \to π/2}\frac{2x\sin(x) - π}{\sin \left(\dfrac\pi2 -x\right)} &\\= &\lim_{x \to π/2}\frac{2x\sin(x) - π}{\left(\dfrac\pi2 -x\right)\dfrac{\sin \left(\dfrac\pi2 -x\right)}{\dfrac\pi2 -x}} &\\=&2\lim_{x \to π/2}\frac{x\sin(x) - \dfracπ2}{\left(\dfrac\pi2 -x\right)}&\\=&2\lim_{x \to π/2}\frac{x\sin(x) - x - \left(\dfracπ2 -x \right)}{\left(\dfrac\pi2 -x\right)}&\\=&-2 +2\lim_{x \to π/2}\frac{x(\sin(x) - 1) }{\left(\dfrac\pi2 -x\right)} &\\=&-2+\pi\lim_{x \to π/2}\frac{\cos\left(\dfrac\pi2 - x\right) - 1 }{\left(\dfrac\pi2 -x\right)}&\\=&-2 +\pi\lim_{x \to π/2}-\frac{\sin^2\left(\dfrac\pi4 - \dfrac x2\right) }{\left(\dfrac\pi2 -x\right)} = -2\end{align}$$

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Substitute $y=\frac\pi 2 - x$. Then

$$\lim_{x \to π/2}\frac{2x\sin x - π}{\cos x} = \lim_{y\to0}\frac{2(\frac\pi 2-y)\cos y - π}{\sin y}$$ and the function under the limit is $$\frac{2(\frac\pi 2-y)\cos y - π}{\sin y} = \frac{(\pi-2y)\cos y - π}{\sin y} \\ = \frac{\pi(\cos y - 1)-2y\cos y}{\sin y} \\ = \pi\frac{\cos y-1}{\sin y}-\frac{2y}{\sin y}\cos y \\ = \pi\frac{\color{red}{\cos y}-\color{blue}1}{\sin y}-2\frac y{\sin y}\cos y \\ = \pi\frac{\color{red}{\cos(2\cdot\frac y2)}-\color{blue}1}{\sin(2\cdot\frac y2)}-2\frac y{\sin y}\cos y \\ = \pi\frac{\color{red}{(\cos^2\frac y2-\sin^2\frac y2)}-\color{blue}{(\sin^2\frac y2+\cos^2\frac y2)}}{2\sin\frac y2\cdot\cos\frac y2}-2\frac y{\sin y}\cos y \\ = \pi\frac{-2\sin^2\frac y2}{2\sin\frac y2\cdot\cos\frac y2}-2\frac y{\sin y}\cos y \\ = -\pi\frac{\sin\frac y2}{\cos\frac y2}-2\frac y{\sin y}\cos y \\ = -\pi\tan\frac y2-2\frac y{\sin y}\cos y $$ We know (...?) that $$\tan 0 = 0$$ $$\cos 0 = 1$$ and $$\lim_{y\to 0}\frac y{\sin y} = 1$$ so the limit sought is $$\lim_{y\to 0} \left(-\pi\tan\frac y2-2\frac y{\sin y}\cos y\right) = -\pi \cdot 0 - 2\cdot 1 \cdot 1 = \boxed{-2}$$

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