Confusion in limits with trigonometric functions Evaluate $$L=\lim_{r\to0} \frac{r\cos r}{r\cos r + \sin r}$$
In the solution it is written that as $r\to0$, $\sin r = r$ and $\cos r = 1$. Hence, we replace the trigonometric functions with $r$ and $1$ so that we can evaluate the limit easily. Therefore,
$L=\lim_{r\to0} \frac{r\cdot1}{r\cdot1 + r}=\frac{1}{2}$.
Now, consider the limit $G =\lim_{r\to0} (\frac{1}{r^{2}}-\frac{1}{\sin^{2}r})$. If we apply the above logic to this question, we will get $G=0$, which is wrong.
So why is the same logic not applicable in the second case?
 A: I believe that the idea is that close enough to $0$, $\sin(r)=r+o(r)$, where $\frac{o(r)}{r}\to 0$ when $r\to 0$ (this is apparent if you calculate the Taylor expansion of $\sin r$). So if we replace $\sin(r)$ with $r+o(r)$ we see that: $$\lim_{r\to 0}\frac{r\cos(r)}{r\cos(r)+r+o(r)}=\lim_{r\to 0}\frac{r\cos(r)}{r(\cos(r)+1+\frac{o(r)}{r})}=\lim_{r\to 0}\frac{\cos(r)}{\cos(r)+1+\frac{o(r)}{r}}=\frac{1}{2}$$
Their argument is not as much "faulty" as it is not formal, because obviously $\sin r$ is not $r$ when $r\to 0$, it's just that $\sin(r)-r$ is $o(r)$, meaning that when taking limits their "contribution" to the limit is the same. They way I calculated the limit puts this in a more formal manner, and it is a routine way to solve limits of this kind.
Another reason they may have wrote that as $r\to 0$, $\sin r=r$ is because $\lim_{r\to 0}\frac{sin(r)}{r}=1$. Either way, it is not a formal argument. If one wants to solve the limit using this argument, it will look like this: $$\lim_{r\to 0}\frac{r\cos(r)}{r\cos(r)+\sin r}=\lim_{r\to 0}\frac{r\cos(r)}{r(\cos(r)+\frac{\sin r}{r})}=\lim_{r\to 0}\frac{\cos(r)}{(\cos(r)+\frac{\sin r}{r})}=\frac{1}{2}$$
A: Let me show you more clearly and precisely that what are you doing in the second case. You are basically doing this :
$\lim_{r\to 0}( \frac{1}{r^2} - \frac{1}{\sin^2r}) $ = $\lim_{r\to 0} \frac{1}{r^2} -$ $\lim_{r\to0}\frac{1}{\sin^2r}$
Since ${\sin r\to r}$ as $r\to 0$ , hence
$$\lim_{r\to 0} \frac{1}{r^2} -\lim_{r\to0}\frac{1}{\sin^2r}=\lim_{r\to 0} \frac{1}{r^2} -\lim_{r\to0}\frac{1}{r^2}$$
Again ,
$$\lim_{r\to 0} \frac{1}{r^2} -\lim_{r\to0}\frac{1}{r^2} = 
\lim_{r\to 0}(\frac{1}{r^2} -\frac{1}{r^2}) = 0$$
You did distribution and re-association of limits which is wrong.
Book author did only distribution and NO re-association of limits which is correct.
This is a very common mistake while studying limits for the first time. Hence, the book solution is correct as he reached the final answer in one step after distribution; while you did more than one step after distribution.
