3
$\begingroup$

Find the value of the limit $$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$$ where $[\cdot]$ denotes the greatest integer function or the box function.

My attempt: I am aware of the standard limits $$\lim_{x\to0}\left(\frac{\sin^{-1}(x)}{x} \right) = 1 $$ and $$\lim_{x\to0}\left(\frac{\tan^{-1}(x)}{x} \right) = 1 $$ but am not sure how will I apply the box function on this limit.

Any detailed explanation to help me understand this concept will be appreciated.

$\endgroup$
  • 2
    $\begingroup$ Do you mean "floor function " ? $\endgroup$ – Khosrotash Jun 6 '17 at 9:52
  • $\begingroup$ Note that your function is even and hence it is sufficient to consider $x\to 0^{+}$. And then you have the inequality $\sin x <x <\tan x$ and limits $(\sin x) /x\to 1,(\tan x) /x\to 1$. Of course you need to convert these inequalities and limits to those involving inverse functions. $\endgroup$ – Paramanand Singh Jun 8 '17 at 0:39
4
$\begingroup$

From

$$\arcsin|x|>|x|$$ and $$\arctan|x|<|x|$$ for small $|x|$, you draw

$$100\frac{\arcsin|x|}{|x|}>100$$ and

$$100\frac{\arctan|x|}{|x|}<100.$$

Then as the limits of these expressions are both $100$ you can find a neighborhood of $0$ where

$$\left\lfloor100\frac{\arcsin|x|}{|x|}\right\rfloor=100$$ and

$$\left\lfloor100\frac{\arctan|x|}{|x|}\right\rfloor=99.$$

$\endgroup$
  • $\begingroup$ can we prove mathematically that in punctured neighbourhood of zero $\frac{\sin^{-1}x}{x} \lt \frac{101}{100}$, because then only we can conclude that limit is $100$ $\endgroup$ – Umesh shankar Jun 15 '18 at 13:48
  • $\begingroup$ The function $\arcsin x/x$ is continuous and tends to $1$. So you can always find a neighborhood such that $\arcsin x/x<1+\epsilon$. $\endgroup$ – Yves Daoust Jun 15 '18 at 13:54
0
$\begingroup$

Your limit is:

$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor + \lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}$$

Split it up:

$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor} + \lim_{x\to0}{\lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}$$

Now, lets consider the left part first:

$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor}$$

While, you're right the limit $\lim_{x\to0}{\left( \frac{\sin^{-1}{x}}{x}\right)}=1$, however the floor function can change that.

Keep in mind that the value of $\sin^{-1}{x}$ is slightly greater than $x$, for $x>0$ and slightly less that $x$, for $x<0$. This would give us:

$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor}=100$$

Desmos® graphing calculator confirms this:

enter image description here

However, for

$$\lim_{x\to0}{\lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}$$

The situation is different. Here, $\tan^{-1}{x}$ is less than $x$ for $x>0$, and greater than $x$, for $x<0$. This means that:

$$\lim_{x\to0}{\lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}=99$$

Again, Desmos® gives us a good reason why:

enter image description here

Combining both results, we get:

$$\lim_{x\to0}{\lfloor 100\frac{\sin^{-1}{x}}{x}\rfloor + \lfloor 100\frac{\tan^{-1}{x}}{x}\rfloor}=199$$

I hope this helps. Leave a comment if you didn't understand any part of the above.

$\endgroup$
  • $\begingroup$ the $\tan^{-1}$ part is wrong, final result should be $199$ $\endgroup$ – stity Jun 6 '17 at 10:20
  • $\begingroup$ You're right, silly of me @stity $\endgroup$ – Pritt Balagopal Jun 6 '17 at 10:21
  • 1
    $\begingroup$ You should have checked with Desmos prior to posting :-) By the way, relying on a grapher can be a bad idea. $\endgroup$ – Yves Daoust Jun 6 '17 at 10:23
  • $\begingroup$ @YvesDaoust I'm on my mobile, it's pretty hard to type accurately. But I'll keep your point in mind. 😊 $\endgroup$ – Pritt Balagopal Jun 6 '17 at 10:25
  • 2
    $\begingroup$ @PrittBalagopal: amateur maths is less risky than amateur chemistry ;-) $\endgroup$ – Yves Daoust Jun 6 '17 at 10:49
-1
$\begingroup$

Since $$\lim_{x\to 0}{\frac{\sin^{-1}(x)}{x}} =1$$ and $\frac{\sin^{-1}(x)}{x}\geq1$ $$\exists \epsilon >0 \ / \ \forall x \in ]-\epsilon,\epsilon[, 1\leq\frac{\sin^{-1}(x)}{x}<1.01 $$ then $$ \forall x \in ]-\epsilon,\epsilon[, 100\leq 100\frac{\sin^{-1}(x)}{x}<101 $$ $$ \forall x \in ]-\epsilon,\epsilon[, \left[ 100\frac{\sin^{-1}(x)}{x}\right] = 100 $$ $$\lim_{x\to 0}{\left[ 100\frac{\sin^{-1}(x)}{x}\right]} = 100$$

Can you apply the same method to $\tan^{-1}$ ? But remember that $\frac{\tan^{-1}(x)}{x}\leq1$

$\endgroup$
  • $\begingroup$ From where you find an upper bound $1.01$ ? $\endgroup$ – Khosrotash Jun 6 '17 at 10:07
  • $\begingroup$ @Khosrotash, that's the definition of a limit : $\lim_{x \to a}{f(x)}=L \iff \forall \delta >0, \exists \epsilon > 0 / \forall x \in ]a-\epsilon, a+\epsilon[, L-\delta \leq f(x) \leq L + \delta$ So one can set $ \delta$ to any value $>0$ and I set it to $0.01$ $\endgroup$ – stity Jun 6 '17 at 10:14
-1
$\begingroup$

When $x\to 0 $ $$\sin x\sim x-\frac{x^3}{6} \space , \space \space\space\sin^{-1} x\sim x+\frac{x^3}{6}\\\tan x\sim x+\frac{x^3}{3}\space ,\space\space \space\tan^{-1} x\sim x-\frac{x^3}{3} $$so $$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]=\\ \lim_{x\to0}\left[100\frac{x+\frac{x^3}{6}}{x}\right]+\left[100\frac{x-\frac{x^3}{3}}{x}\right]=\\ \lim_{x\to0}\left[100(1+\frac{x^2}{6})\right]+\left[100(1-\frac{x^2}{3})\right]=\\ \lim_{x\to0}\left[100+\frac{100x^2}{6}\right]+\left[100-\frac{100x^2}{3})\right]=\\ \left[100^+\right]+\left[100^-\right]=100+99$$

$\endgroup$
  • $\begingroup$ The result is indeed correct, but for rigor, you should show that the remainder of the Tayor development can be neglected. $\endgroup$ – Yves Daoust Jun 6 '17 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.