Inverse Trig Limits

Calculus Level 5

$\large \lim_{x \to 0} \left( \left \lfloor 100\ \frac{x}{\sin^{-1}(x)} \right \rfloor +\left \lfloor 100\ \frac{\tan^{-1}(x)}{x} \right \rfloor \right) =\ ?$

None of these choices 200 199 198

4 solutions

Shashwat Shukla
Jan 30, 2015

A purely analytic solution can be as follows:

The Taylor series for $sin^{-1}x$ and $tan^{-1}x$ centered at zero are:

$sin^{-1}x=x+\frac{1}{2}\frac{x^3}{3}+\frac{1}{2}\frac{3}{4}\frac{x^5}{5}...$

Clearly $\frac{x}{sin^{-1}x}<1$ in the vicinity of 0.

Similarly, $tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}...$ Clearly $\frac{tan^{-1}x}{x}<1$ in the vicinity of 0.

The required result follows.

why 199 can't be the answer?

Lipsa Kar - 5 years, 3 months ago

This seems to me to be a trick question. At x = 0, both the arcsin x and arctan x are also 0. This results in both terms inside the floor brackets becoming 100 x 0/0. If we apply l'Hospital's Rule to both terms, the resulting expressions for both evaluate unambiguously to 1 at x = 0. One hundred times 1 is exactly 100, and the greatest integer function of 100 is 100, yielding a sum of 200. But we are asked to believe that the terms do not reach 1, because x never identically equals 0. However, since we know by other means that the limit does exist and it is 1, we can get arbitrarily close to 1 if x gets arbitrarily close to 0. In other words, the two terms are equal to .0.99999... and 100 times that is 99.9999...The greatest integer function is then 99 for each, and the sum is 198. Here's the catch. if we approach infinitesimally close to 0, the result is 0.9999999....... to infinity. It is trivial to demonstrate that, although 0.9999999.... ad inf. is less than 1 at any truncated version, it is actually identically equal to 1.0 in the repeating decimal form. So, if 0.9999...ad inf = 1.0, then 99.99999...ad inf = 100.0. The greatest integer function seems to be different, but it is just as easy to demonstrate that 99.9999...ad inf = 100.0, and the correct answer is 200.

Tom Capizzi - 4 years, 11 months ago

Tanishq Varshney
Jan 29, 2015

$\displaystyle \lim_ {x\to 0} [\frac{x}{\sin^{-1} x}]$ = 0

$\displaystyle \lim _{x\to 0} [\frac{\tan^{-1} x}{x}]$ = 0

where[.] represents greatest integer function which also means that they have value less than 1 .

so $[100 \times \text valueslightly less than 1]=99$ where [.] is greatest integer fuction

.01 is less than 1, so you seem to be saying that [100x.01] = 99

But, [100x.01] = 1

Or does "valuelessthan1" mean slightly less than 1?

I don't follow your solution.

More importantly, $\lim _{ x\rightarrow 0 }{ \frac { x }{ arcsin\quad x } } =1$ It is NOT 0.

Danny Whittaker - 6 years, 4 months ago

By "valuelessthan1", I am quite sure he meant slightly less than 1. And the limit does equal 0 when it is a limit of greatest integer of that function.

Dhruva Patil - 6 years, 4 months ago

Duh, thanks. I was doing greatest integer of limit instead of limit of greatest integer. That's what happens when I try to think after a long day of teaching!

Danny Whittaker - 6 years, 4 months ago

Gautam Sharma
Jan 30, 2015

Line $y=x$ passes above the the graph of $\tan ^{ -1 }{ x }$ and they cut each other only at origin.

$\lim _{ x\rightarrow 0 }[{ \frac { \tan ^{ -1 }{ x } }{ x } }] =0$

similarily graph of $\sin ^{ -1 }{ x }$ passes above the line $y=x$ and they cut each other only at origin.

$\lim _{ x\rightarrow 0 }[{ \frac { x }{ \sin ^{ -1 }{ x } } }] =0$

rest figure out yourself.

Mohamed Laghlal
Jan 30, 2015

Just note that $\frac{x}{\arcsin x}$ and $\frac{arctan x}{x}$ tends to 1 when $x$ tends to 0, but they are never 1.

Inverse Trig Limits

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