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

Question

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 .

Answer 1

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

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:

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

Answer 2

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

Answer 3

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 $2\cos^2(y+\frac{\pi}{2})$.

Answer 4

Use the relevant duplication formula: $$\frac{\sin x-1}{1+\cos 2x}=\frac{\sin x-1}{2\cos^2x}=\frac{\sin^2x-1}{2\cos^2x(\sin x+1)}=\frac{-\cos^2x}{2\cos^2x(\sin x+1)}=-\frac 1{2(1+\sin x)}\to-\frac14.$$

Answer 5

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

Answer 6

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*}

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

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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!