Will the answer change?

Calculus Level 5

$\large \lim_{n \to \infty} \left | \cos\left( \dfrac{\pi \sqrt{n^2+n}}{2} \right) \right |$

What is the value of the limit above for integers $n$ ? Give your answer up to 3 decimal places.

Submit $12345$ as the answer if the limit does not exist.

Title of the problem given in reference to the problem : limits

The answer is 0.707.

2 solutions

Ayush Verma
Apr 24, 2015

${ lim }_{ n\rightarrow \infty }\left| \cos { \left( \cfrac { \pi \sqrt { { n }^{ 2 }+n } }{ 2 } \right) } \right| =?\\ \\ but,\quad \sqrt { { n }^{ 2 }+n } =n{ \left( 1+\cfrac { 1 }{ { n } } \right) }^{ 1/2 }\\ \\ \sqrt { { n }^{ 2 }+n } =n\left( 1+\cfrac { 1 }{ { 2n } } -\cfrac { 1 }{ 8{ n }^{ 2 } } +... \right) =\left( n+\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 8n } +... \right) \\ \\ so\quad for\quad n\rightarrow \infty ,\quad \sqrt { { n }^{ 2 }+n } \quad will\quad be\quad almost\quad \quad \left( n+\cfrac { 1 }{ 2 } \right) \\ \\ \therefore \quad { lim }_{ n\rightarrow \infty }\left| \cos { \left( \cfrac { \pi \sqrt { { n }^{ 2 }+n } }{ 2 } \right) } \right| ={ lim }_{ n\rightarrow \infty }\left| \cos { \left( \cfrac { \pi \left( n+\cfrac { 1 }{ 2 } \right) }{ 2 } \right) } \right| \\ \\ ={ lim }_{ n\rightarrow \infty }\left| \cos { \left( \cfrac { n\pi }{ 2 } +\cfrac { \pi }{ 4 } \right) } \right| ={ lim }_{ n\rightarrow \infty }\left| \pm \sin { \cfrac { \pi }{ 4 } } \right| =\cfrac { 1 }{ \sqrt { 2 } } =0.707$

Moderator note:

For clarity, you should elaborate on why you pick only the first two terms of the expansion $\left ( 1 + \frac 1 n \right )^{\frac 1 2 }$ .

I think we can also do this by binomial approximation.Since n tends to infinity therefore 1/n tends to 0.Therefore,(1+1/n)^1/2 = 1+1/2n and we can proceed further without using multinomial theorem i guess.

Swayam Prakash Kar - 8 months, 3 weeks ago

Priyesh Pandey
Apr 28, 2015

How could it be on 5th lvl

Will the answer change?

Title of the problem given in reference to the problem : limits

The answer is 0.707.

2 solutions

Moderator note:

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