Limit-o-saurus

Calculus Level 3

Let x 1 , x 2 , x 3 , . . . x n x_1,x_2,x_3,...x_n be real numbers such that x 1 > x 2 > x 3 . . . . > x n x_1> x_2 > x_3....>x_n .

Also x 1 = tan ⁡ − 1 2 x_1=\tan^{-1} {2} and sin ⁡ ( x n + 1 − x n ) + 1 2 n + 1 sin ⁡ ( x n ) sin ⁡ ( x n + 1 ) = 0 \sin (x_{n+1}-x_n)+\dfrac{1}{2^{n+1}} \sin(x_n) \sin(x_{n+1})=0 .

Let I = lim ⁡ n → ∞ cot ⁡ x n I= \lim _{ n\rightarrow \infty }{ \cot { x_n} }

Evaluate [ I ] \large [ {I}] .

Note : [ ] represents the greatest integer function.


The answer is 1.

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Vijay Raghavan by Vijay Raghavan , 24, India

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2 solutions

Vijay Raghavan
Feb 25, 2014

Given that

sin ⁡ ( x n + 1 − x n ) = − 1 2 n + 1 sin ⁡ ( x n ) s i n ( x n + 1 ) \sin(x_{n+1}-x_{n})=\dfrac{-1}{2^{n+1}} \sin(x_n)\ sin(x_{n+1})

Which gives us:

cot ⁡ ( x n + 1 ) − cot ⁡ ( x n ) = 1 2 n + 1 \cot(x_{n+1})-\cot(x_{n})=\dfrac{1}{2^{n+1} }

This series telescopes :

cot ⁡ ( x 2 ) − cot ⁡ ( x 1 ) = 1 2 2 \cot(x_2)-\cot(x_1) =\dfrac{1}{2^2}

cot ⁡ ( x 3 ) − cot ⁡ ( x 2 ) = 1 2 3 \cot(x_3)-\cot(x_2) =\dfrac{1}{2^3}

.....

cot ⁡ ( x n + 1 ) − cot ⁡ ( x n ) = 1 2 n + 1 \cot(x_{n+1})-\cot(x_n) =\dfrac{1}{2^{n+1}}

Also, we have x 1 = tan ⁡ − 1 2 ⟹ c o t x 1 = 1 2 x_1=\tan^{-1}2 \implies cot{x_1}=\dfrac{1}{2}

We have

cot ⁡ ( x n + 1 ) = 1 2 + 1 2 2 + 1 2 3 + 1 2 4 . . . + 1 2 n + 1 \cot(x_{n+1})=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}...+\dfrac{1}{2^{n+1}}

⟹ cot ⁡ ( x n ) = 1 2 + 1 2 2 + 1 2 3 + 1 2 4 . . . + 1 2 n \implies \cot(x_n)=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}...+\dfrac{1}{2^{n}}

cot ⁡ ( x n ) = 1 2 ( 1 − 1 2 n ) 1 − 1 2 = 1 − 1 2 n \cot(x_n)=\dfrac{\dfrac{1}{2}(1-\dfrac{1}{2^n})}{1-\dfrac{1}{2}}= 1-\dfrac{1}{2^n}

I = lim ⁡ n → ∞ cot ⁡ ( x n ) = 1 I= \lim_{n \to \infty} \cot(x_n) =\boxed{1}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

This problem actually came in my AITS examination..

Anish Puthuraya - 7 years, 3 months ago

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Oh. didn't know that. This one came in my test too.

Vijay Raghavan - 7 years, 3 months ago

Does the paper contain more questions like this? I would definitely like to try them. If it is ok for you, can you tell me which AITS you are talking about?

Thanks! :)

Pranav Arora - 7 years, 3 months ago

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I am studying at IITian's PACE, so this is from the AITS-7 Mains examination of PACE coaching institute.

Actually this was the only problem that bothered me at the time of the exam, the other problems were quite easy (well, too easy for a brilliant mind as you)

I secured an All India Rank of 69 in that examination (well, I've had better days.)

Anish Puthuraya - 7 years, 3 months ago

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@Anish Puthuraya – Thanks Anish! And that's a pretty good score in my opinion, keep it up. :)

Btw, if you have time, can you post a solution to this: Definite Integral . I am looking for alternative approaches to this problem. :)

Best of luck with your boards!

Pranav Arora - 7 years, 3 months ago

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@Pranav Arora – No, I solved it by using the same method..But, Ill try and figure out an alternative.

Best of luck to you as well!

Anish Puthuraya - 7 years, 3 months ago

yeah, i remember this one

Mandar Sohoni - 7 years, 3 months ago

yes, i agree that I = 1-(1/2)^n and hence when limit n tends to infinity I = 1 but you r question asks for [I] which greatest integer function. hence no matter how big may 'n' be it is actually be going to be less than 1 always hence the answer should be zero... and not 1

Vishal Sharma - 7 years, 2 months ago

Mistaken sorry

Vishal Ajwani - 7 years, 2 months ago

Don't you think greatest integer function would give 0 as the answer?

Vishal Ajwani - 7 years, 2 months ago
Apoorv Pandey
Mar 6, 2014

vijay the answer you have given is wrong as 0.5+0.25+0.125+... will approach 1 from below and the integer part of I will be 0

1 Helpful 0 Interesting 0 Brilliant 0 Confused

The limit is exactly equal to 1 1 . The integer part of it too will equal one.

Parth Thakkar - 7 years, 3 months ago

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