5
$\begingroup$

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?

$\endgroup$
10
  • 2
    $\begingroup$ Nice question! What your example actually proves is that the solution that you were given is not correct. It reaches the right answer with a faulty argument. $\endgroup$ Jan 1, 2023 at 11:44
  • $\begingroup$ @JoséCarlosSantos I see. So basically we cannot replace functions inside the limit this way, right? I suppose taking $r$ outside and canceling is the best way to solve the 1st limit. $\endgroup$
    – Sasikuttan
    Jan 1, 2023 at 11:55
  • $\begingroup$ I see nothing wrong with the reasoning of the given solution (poor notations set aside, see here). The reason why it does not apply to your example is that you cannot substract (or add) equivalents because of their definition. $\endgroup$
    – nicomezi
    Jan 1, 2023 at 12:10
  • 2
    $\begingroup$ I would do\begin{align}\lim_{r\to0}\frac{r\cos r}{r\cos r+\sin r}&=\lim_{r\to0}\frac{\cos r}{\cos r+\frac{\sin r}r}\\&=\frac{\lim_{r\to0}\cos r}{\lim_{r\to0}\cos r+\frac{\sin r}r}\\&=\frac1{1+1}\\&=\frac12.\end{align} $\endgroup$ Jan 1, 2023 at 12:14
  • 1
    $\begingroup$ @Curiouserandcuriouser Because $g(x)$ and $f(x)$ are said equivalent at $a$ if $\lim f(x)/g(x) \to 1$ as $x \to a$. Assume $f_1,f_2$ are equivalent to $g_1,g_2$, we have that $\underset{x \to a }\lim f_1 / f_2 = \underset{x \to a }\lim f_1 g_1 g_2 / (f_2 g_1 g_2) = \underset{x \to a }\lim g_1 / g_2 $. If one rather evaluates $\underset{x \to a } \lim f_1 + f_2$, you might have problem with indeterminate forms as in the example you gave (but it will work if all limits are finite). $\endgroup$
    – nicomezi
    Jan 1, 2023 at 12:31

2 Answers 2

3
$\begingroup$

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: $$\lim_{r\to 0}\frac{r\cos(r)}{r\cos(r)+\sin r}=\lim_{r\to 0}\frac{r\cos(r)}{r(\cos(r)+\frac{\sin r}{r})}=\lim_{r\to 0}\frac{\cos(r)}{(\cos(r)+\frac{\sin r}{r})}=\frac{1}{2}$$

$\endgroup$
1
  • 2
    $\begingroup$ I agree with this answer and would like to add that a correct notation woud be $\sin r \underset{r \to 0}\sim r$. $\endgroup$
    – nicomezi
    Jan 1, 2023 at 12:14
2
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ I don't agree that that's what the OP did. Instead, it was simply$$\lim_{r\to0}\left(\frac1{r^2}-\frac1{\sin^2r}\right)=\lim_{r\to0}\left(\frac1{r^2}-\frac1{r^2}\right)=0.$$ $\endgroup$ Jan 1, 2023 at 15:24
  • 1
    $\begingroup$ @JoséCarlosSantos The OP's step from the LHS to RHS is the thing that I have explained more clearly in starting of my answer . It is not possible to change $\sin r$ to $r$ without knowing that the limit goes to zero. This is what I have done in the starting of my answer . $\endgroup$ Jan 1, 2023 at 15:27
  • $\begingroup$ The OP has used the fact(knowledge) that $r$ goes to zero. And in mathematical terms, it is same as distributing limits in this case. $\endgroup$ Jan 1, 2023 at 15:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .