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How do I evaluate this: $$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)}$$? without using L'Hospital's rule.

Attempt : I used trigonomeitric transformation and I used Taylor series I got this : $$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)} = \lim_{y\to 0}\displaystyle \frac {\cos y-cos²y}{2\cos² (2y+\pi)}-\frac{1}{2}$$ where $y=x-\frac{\pi}{2}$, but by this way i got : $$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)}=\frac{-1}{2}$$ and by L'Hospital's rule is :$\frac{-1}{4}$ Then where is my problem if I'm true

Thank you for any help .

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$$\lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos 2x)} = \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos^2x-\sin^2x)} = \\ = \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+(1-\sin^2x)-\sin^2x)} = \\ = \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{2(1-\sin^2x)} = \\ = \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{2(1-\sin x)(1+\sin x)} = \\ = \lim_{x\to \frac{\pi}{2}} \frac {-1}{2(1+\sin x)} = -\frac{1}{4}\\$$

Using Taylor for $x_0 = \frac{\pi}{2}$...

$$\sin(x) = 1 - (\pi/2 - x)^2/2 + \ldots$$ $$\cos(2x) = -1 + 2(\pi/2 - x)^2 + \ldots$$ Then:

$$\lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos 2x)} = \lim_{x\to \frac{\pi}{2}} \frac { 1 - (\pi/2 - x)^2/2-1+ \ldots}{(1-1 + 2(\pi/2 - x)^2 + \ldots)} = \\ =\lim_{x\to \frac{\pi}{2}} \frac {- (\pi/2 - x)^2/2+ \ldots}{2(\pi/2 - x)^2 + \ldots} = -\frac{1}{4}$$

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  • $\begingroup$ in any way thank you for this answer , hop to show me where is the problem in my attempt using taylor series ? $\endgroup$ – zeraoulia rafik Sep 21 '15 at 20:11
  • $\begingroup$ @zeraouliarafik here it is. You were evaluating wrongly the Taylor series... $\endgroup$ – the_candyman Sep 21 '15 at 20:21
  • $\begingroup$ Ok, thank you i got it now $\endgroup$ – zeraoulia rafik Sep 21 '15 at 20:22
  • $\begingroup$ @zeraouliarafik you are welcome $\endgroup$ – the_candyman Sep 21 '15 at 20:26
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Use the relevant duplication formula: $$\frac{\sin x-1}{1+\cos 2x}=\frac{\sin x-1}{2\cos^2x}=\frac{\sin^2x-1}{2\cos^2x(\sin x+1)}=\frac{-\cos^2x}{2\cos^2x(\sin x+1)}=-\frac 1{2(1+\sin x)}\to-\frac14.$$

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Given $$\displaystyle \lim_{x\rightarrow \frac{\pi}{2}}\frac{\sin x-1}{1+\cos 2x}\;,$$ Now Put $\displaystyle x=\frac{\pi}{2}-y\;,$ Then when $\displaystyle x\rightarrow \frac{\pi}{2}\;,$ Then $y\rightarrow 0$

So we get $$\displaystyle\lim_{y\rightarrow 0}\frac{\sin \left(\frac{\pi}{2}-y\right)-1}{1+\cos\left(\pi-2y\right)} = \lim_{y\rightarrow 0}\frac{\cos y-1}{1-\cos 2y} = -\lim_{y\rightarrow 0}\frac{1-\cos y}{2\sin^2 y}$$

So we get $$\displaystyle -\lim_{y\rightarrow 0}\frac{1-\cos y}{2\sin^2 y}\times \frac{1+\cos y}{1+\cos y} = -\lim_{y\rightarrow 0}\frac{\sin^2y}{2\sin^2 y}\times \lim_{y\rightarrow 0}\frac{1}{1+\cos y} = -\frac{1}{2}\times\frac{1}{2} = -\frac{1}{4}$$

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  • $\begingroup$ nice answer , thank you very much for this $\endgroup$ – zeraoulia rafik Sep 21 '15 at 20:17
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Considering the identity $\cos(2x) = 1-2\sin^2 x$, we have:$$\frac{\sin x-1}{1+\cos(2x)}=\frac{\sin x -1}{2-2\sin^2x}=-\frac{1-\sin x}{2(1-\sin x)(1+\sin x)}.$$

Thus the given limit is: $$\lim_{x\to \pi/2}\frac{\sin x-1}{1+\cos (2x)}=\lim_{x\to\pi/2}-\frac{1}{2(1+\sin x)}=-\frac 14$$

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Lemma: If $f(x), g(x)$ are $C^2$ in a neighborhood of $x=c$ and $f(c) = g(c) = 0$, then $$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}.$$

Proof: For $x$ in a neighborhood of $c$, $$f(c) = f(x) + f'(x)(c-x) + \frac{1}{2}f''(\epsilon_f[x,c])(c-x)^2$$ and $$g(c) = g(x) + g'(x)(c-x) + \frac{1}{2}g''(\epsilon_g[x,c])(c-x)^2,$$ where $\epsilon_f(x,c)$ is between $c$ and $x$. Then \begin{align*} \lim_{x\to c} \frac{f(x)}{g(x)} &= \lim_{x\to c} \frac{f(c) - f'(x)(c-x) - \frac{1}{2}f''(\epsilon_f[x,c])(c-x)^2}{g(c) - g'(x)(c-x) - \frac{1}{2}g''(\epsilon_g[x,c])(c-x)^2}\\ &= \lim_{x\to c} \frac{0 - f'(x)(c-x) - \frac{1}{2}f''(\epsilon_f[x,c])(c-x)^2}{0 - g'(x)(c-x) - \frac{1}{2}g''(\epsilon_g[x,c])(c-x)^2}\\ &= \lim_{x\to c} \frac{f'(x) + \frac{1}{2}f''(\epsilon_f[x,c])(c-x)}{g'(x) + \frac{1}{2}g''(\epsilon_g[x,c])(c-x)}\\ &= \lim_{x\to c} \frac{f'(x)}{g'(x)}. \end{align*}


Now for the main problem: apply the above lemma ("user7530's rule") twice to get: $$\lim_{x\to \pi/2} \frac{\sin x - 1}{1+\cos(2x)} = \lim_{x\to \pi/2} \frac{\cos x}{-2\sin(2x)} = \lim_{x\to \pi/2} \frac{-\sin x}{-4\cos 2x} = -\frac{1}{4}.$$


Yes, this is a "joke" answer to prove a point. And yet it is still insufficient to convey the extent to which I despise "without l'Hopital" questions, especially in cases like this one where everything in sight is smooth and analytic and there is no conceivable reason to avoid it. People wouldn't think "without l'Hopital" is an interesting restriction if they truly understand how and why it worked. It's especially ridiculous when some of the "without l'Hopital" answers then go on and use Taylor's theorem as part of the solution!

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You need neither L'Hospital nor Taylor: $\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{\sin x-1}{1+\cos(2x)}=-\lim_{x\to\frac{\pi}{2}}\frac{1-\sin x}{2\cos^2x}=-\frac{1}{2}\lim_{x\to\frac{\pi}{2}}\frac{(\sin\frac{x}{2}-\cos\frac{x}{2})^2}{[(\cos\frac{x}{2}-\sin\frac{x}{2})(\cos\frac{x}{2}+\sin\frac{x}{2})]^2}$$=-\frac{1}{2}\cdot\frac{1}{(\cos\frac{\pi}{4}+\sin\frac{\pi}{4})^2}=-\frac{1}{4}$

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In step where you changed $1+\cos(2x)$ to $2\sin^2x$ you still multiplied 2 into x when you changed $x=y+\frac{\pi}{2}$. and computed $2\cos^2(2y+\pi)$ but its $2\cos^2(y+\frac{\pi}{2})$.

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