How to compute $\lim\limits_{\theta \to \frac{\pi}{2}^+} \frac{\cos{\theta}-2\sin{\theta}\cos{\theta}}{\sin^2{\theta}-\cos^2{\theta}-\sin{\theta}}$?

Question

How can we compute the limit

$$\lim\limits_{\theta \to \frac{\pi}{2}^+} \frac{\cos{\theta}-2\sin{\theta}\cos{\theta}}{\sin^2{\theta}-\cos^2{\theta}-\sin{\theta}}\tag{1}$$

Here is context on why I want to solve this limit.

Consider the function $f(\theta)=1-\sin{\theta}$.

If $(r,\theta)$ are polar coordinates, then $$r=f(\theta)\tag{2}$$ represents the following curve in polar coordinates (it's called a cardioid)

Now suppose we want to verify that the tangent at $x=0$ is vertical.

Note that $r=0 \implies 1-\sin{\theta}=0 \implies \theta=\frac{\pi}{2}$. That is, The point $(0,0)$ on the graph corresponds to $(0,\pi/2)$ in polar coordinates.

The parametric equations for this curve are

$$(x(\theta),y(\theta))= ((1-\sin{\theta})\cos{\theta},(1-\sin{\theta})\sin{\theta})$$

If $x'(\theta) \neq 0$ we have

$$y'(x)=\frac{y'(\theta)}{x'(\theta)}=\frac{\cos{\theta}-2\sin{\theta}\cos{\theta}}{\sin^2{\theta}-\cos^2{\theta}-\sin{\theta}}\tag{3}$$

However, $x'(\frac{\pi}{2})=0$.

I'd like to compute the limit of $(3)$ as $\theta \to \left ( \frac{\pi}{2} \right ) ^+$ and show that it is $-\infty$.

Answer 1

$$\lim_{\theta\to \pi/2^+}\frac{\cos\theta-\sin(2\theta)}{-\cos(2\theta)-\sin\theta}\stackrel{\theta\mapsto\varphi+\pi/2}{=}\lim_{\varphi\to 0^+}\frac{-\sin\varphi+\sin(2\varphi)}{\cos(2\varphi)-\cos\varphi}=\lim_{\varphi\to 0^+}\frac{\varphi+O(\varphi^3)}{-\frac{3}{2}\varphi^2+O(\varphi^4)}=-\infty. $$

Answer 2

Here is a slightly different one from Jack's answer that does not use the double angle formula.

The Maclaurin series is easier to use (and write) than the Taylor series at $\pi/2$.

By the trigonometric identities $$ \cos(y+\frac\pi2)=-\sin(y),\quad \sin(y+\frac\pi2)=\cos(y) $$ one can write, setting $x=y+\frac\pi2$, $$ \frac{\cos (x)-2\sin(x)\cos(x)}{\sin^2(x)-\cos^2(x)-\sin(x)} =\frac{-\sin(y)+2\sin(y)\cos(y)}{\cos^2(y)-\cos(y)-\sin^2(y)}\tag{1} $$ Now if one substitute $$\sin(y)=y+O(y^2),\quad\cos(y)=1-\frac{y^2}{2}+O(y^4)$$ into (1), then $$ \frac{-\sin(y)+2\sin(y)\cos(y)}{\cos^2(y)-\cos(y)-\sin^2(y)} =\frac{-y+2y+O(y^2)}{(-1-\frac12)y^2+O(y^4)}\to -\infty $$ as $y\to 0^+$.

Answer 3

$$\lim\limits_{\theta \to \frac{\pi}{2}^+} \frac{\cos{\theta}-2\sin{\theta}\cos{\theta}}{\sin^2{\theta}-\cos^2{\theta}-\sin{\theta}}=\lim\limits_{\theta \to \frac{\pi}{2}^+} \frac{\cos{\theta}(1-2\sin{\theta})}{\sin^2{\theta}-1+\sin^2{\theta}-\sin{\theta}}=$$ $$=\lim\limits_{\theta \to \frac{\pi}{2}^+} \frac{\cos{\theta}(1-2\sin{\theta})}{(\sin{\theta}-1)(\sin{\theta}+1+\sin{\theta})}=\lim\limits_{\theta \to \frac{\pi}{2}^+} \frac{\sqrt{1+\sin{\theta}} \cdot (1-2\sin{\theta})}{\sqrt{1-\sin{\theta}}\cdot (2\sin{\theta}+1)}=-\infty$$

Answer 4

If you make $\theta=x+\frac \pi 2$ you are looking for $$-\lim_{x\to 0^+}\cot \left(\frac{3 }{2}x\right)$$

For once, I am not using Taylor series.