Prove that $\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$ doesn't exist without L'hopital rule I need to prove that the following limit doesn't exist:
$$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$$
Itried to simplify $1+\cos(2x)$ into $2\cos^2(x)$, in order to calculate both of the one-sided limits, but I am not sure how to calculate it without using L'hopital rule.
 A: $\cos (3x)=4\cos^{3}x-3\cos x$ and $1+\cos (2x)=2\cos^{2}x$. So the given limit is $\lim_{x \to \pi/2} [2\cos x-\frac 3 {2\cos x}]$ which does not exist.
A: To prove that the limit doesnot exist we proceed as following:
$$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$$
Put $x = \frac{\pi}{2}$ to get
$$ \frac{\cos(\frac{3\pi}{2})}{1+\cos(2\frac{\pi}{2})} = \frac{0}{0} $$ which is the indeterminate form. Now to prove that the limit doesnot exist we consider the Left Hand Limit and the Right Hand Limit for the function.
Left hand limit:
Let $x$ approach $\frac{\pi}{2}$ from a value slightly lower than it i.e. $x \lt \frac{\pi}{2}$. This is the intuition for LHL. Formally,
$$\lim_\limits{x\to {\frac{\pi}{2}}^-}\frac{\cos(3x)}{1+\cos(2x)}$$
Put, $$x = \frac{\pi}{2} - h$$ where $h$ is an infinitely small positive number. Now as $$x = \frac{\pi}{2}-h$$ $x$ tends to $\frac{\pi}{2}$,  $h$ tends to $0$ and the limit becomes:
$$\lim_\limits{x\to {\frac{\pi}{2}^-}}\frac{\cos(3x)}{1+\cos(2x)} = \lim_\limits{h \to 0}\frac{\cos3(\frac{\pi}{2}-h)}{1+\cos2(\frac{\pi}{2}-h)}$$
$$=\lim_\limits{h \to 0}\frac{-\sin3h}{1-\cos2h}$$
Solve it further to get,
$$\lim_\limits{h \to 0}\frac{-\sin3h}{1-\cos2h} = \lim_\limits{h \to 0}\frac{-2\sin(\frac{3h}{2})\cos(\frac{3h}{2})}{2\sin^2{\frac{h}{2}}} = \lim_\limits{h \to 0}\frac{-3}{\sin\frac{h}{2}} = \frac{-3}{0} = -\infty$$
PS: Take care of the negative sign.
Proceed the same for the RHL:
Right hand limit:
Let $x$ approach $\frac{\pi}{2}$ from a value slightly above than it i.e. $x \gt \frac{\pi}{2}$ $$\lim_\limits{x\to {\frac{\pi}{2}}^+}\frac{\cos(3x)}{1+\cos(2x)}$$
Put, $$x = \frac{\pi}{2} + h$$ where $h$ is an infinitely small positive number. Now as $$x = \frac{\pi}{2}+h$$ $x$ tends to $\frac{\pi}{2}$,  $h$ tends to $0$ and the limit becomes:
$$\lim_\limits{x\to {\frac{\pi}{2}^+}}\frac{\cos(3x)}{1+\cos(2x)} = \lim_\limits{h \to 0}\frac{\cos3(\frac{\pi}{2}+h)}{1+\cos2(\frac{\pi}{2}+h)}$$
$$=\lim_\limits{h \to 0}\frac{\sin3h}{1-\cos2h}$$
Solve it further to get,
$$\lim_\limits{h \to 0}\frac{\sin3h}{1-\cos2h} = \lim_\limits{h \to 0}\frac{2\sin(\frac{3h}{2})\cos(\frac{3h}{2})}{2\sin^2{\frac{h}{2}}} = \lim_\limits{h \to 0}\frac{3}{\sin\frac{h}{2}} = \frac{3}{0} = \infty$$
You can see from the above analysis that,
$$LHL = -\infty$$
$$RHL = \infty$$
$$LHL \neq RHL$$
So the limit doesnot exist. Hope this helps...
A: Starting from your last line:
$$L=\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}=\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{2\cos^2 (x)}$$
$$L=\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{2\sin^2 (\pi/2-x)}$$
Change variable $x=\dfrac {\pi}2
-u$:
$$L=\lim_\limits{u\to 0}\frac{\cos(3\pi /2-3u)}{2\sin^2 (u)}=\lim_\limits{u\to 0}\frac{\sin(-3u)}{2\sin^2 (u)}$$
$$L=-\dfrac 12\lim_\limits{u\to 0}\dfrac{\sin (3u)}{3u}\dfrac{u^2}{\sin^2 (u)}\frac 3 u$$
Then use: $$\lim_\limits{u\to 0}\frac{\sin u}{u}=1$$
A: $$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}=\lim_\limits{y\to 0}\frac{\sin(3y)}{1-\cos(2y)}$$ Using Taylor series and long division
$$\frac{\sin(3y)}{1-\cos(2y)}=\frac{3}{2 y}-\frac{7 y}{4}+O\left(y^3\right)$$
