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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.

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    $\begingroup$ Do you mean "floor function " ? $\endgroup$
    – Khosrotash
    Commented Jun 6, 2017 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
    Commented Jun 8, 2017 at 0:39

4 Answers 4

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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

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

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  • $\begingroup$ 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 $100$ $\endgroup$ Commented Jun 15, 2018 at 13:48
  • $\begingroup$ The function $\arcsin x/x$ is continuous and tends to $1$. So you can always find a neighborhood such that $\arcsin x/x<1+\epsilon$. $\endgroup$
    – user65203
    Commented Jun 15, 2018 at 13:54
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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.

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  • $\begingroup$ the $\tan^{-1}$ part is wrong, final result should be $199$ $\endgroup$
    – stity
    Commented Jun 6, 2017 at 10:20
  • $\begingroup$ You're right, silly of me @stity $\endgroup$ Commented Jun 6, 2017 at 10:21
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    $\begingroup$ You should have checked with Desmos prior to posting :-) By the way, relying on a grapher can be a bad idea. $\endgroup$
    – user65203
    Commented Jun 6, 2017 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. 😊 $\endgroup$ Commented Jun 6, 2017 at 10:25
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    $\begingroup$ @PrittBalagopal: amateur maths is less risky than amateur chemistry ;-) $\endgroup$
    – user65203
    Commented Jun 6, 2017 at 10:49
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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 $\frac{\tan^{-1}(x)}{x}\leq1$

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  • $\begingroup$ From where you find an upper bound $1.01$ ? $\endgroup$
    – Khosrotash
    Commented Jun 6, 2017 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
    Commented Jun 6, 2017 at 10:14
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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 $$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]=\\ \lim_{x\to0}\left[100\frac{x+\frac{x^3}{6}}{x}\right]+\left[100\frac{x-\frac{x^3}{3}}{x}\right]=\\ \lim_{x\to0}\left[100(1+\frac{x^2}{6})\right]+\left[100(1-\frac{x^2}{3})\right]=\\ \lim_{x\to0}\left[100+\frac{100x^2}{6}\right]+\left[100-\frac{100x^2}{3})\right]=\\ \left[100^+\right]+\left[100^-\right]=100+99$$

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    $\begingroup$ The result is indeed correct, but for rigor, you should show that the remainder of the Tayor development can be neglected. $\endgroup$
    – user65203
    Commented Jun 6, 2017 at 10:19

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