infinity limits are interesting...

Calculus Level 5

lim ⁡ n → ∞ n sin ⁡ ( 2 π 1 + n 2 ) = ? \large \lim_{n \rightarrow \infty} n\sin \left(2\pi \sqrt{1+n^{2}}\right)= \ ?
where n n is a natural number.

Try for some more interesting problems of Limits and Derivatives.
0 does not exist π \pi 1

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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2 solutions

Prakhar Gupta
Jan 12, 2015

This is a bit tricky question. But it can be solved. The question is :- lim ⁡ n → ∞ n . sin ⁡ ( 2 π 1 + n 2 ) \lim_{n\to\infty} n.\sin (2\pi \sqrt{1+n^{2}}) Here comes the tricky part. We know that sin ⁡ x = sin ⁡ ( x − 2 n π ) \sin x= \sin(x-2n\pi) . We are going to use this in the question. = lim ⁡ n → ∞ n . sin ⁡ ( 2 π 1 + n 2 − 2 n π ) =\lim_{n\to\infty} n.\sin (2\pi\sqrt{1+n^{2}} - 2n\pi) = lim ⁡ n → ∞ n . sin ⁡ ( 2 π ( 1 + n 2 − n ) ) =\lim_{n\to\infty} n.\sin(2\pi(\sqrt{1+n^{2}}-n)) Rationalizing the part inside sin. = lim ⁡ n → ∞ n . sin ⁡ ( 2 π ( 1 1 + n 2 + n ) ) =\lim_{n\to\infty} n.\sin(2\pi(\dfrac{1}{\sqrt{1+n^{2}}+n})) = lim ⁡ n → ∞ n . sin ⁡ ( 2 π n 1 1 n 2 + 1 + 1 ) =\lim_{n\to\infty} n.\sin(\dfrac{2\pi}{n} \dfrac{1}{\sqrt{\dfrac{1}{n^{2}}+1}+1}) Neglecting 1 n 2 \frac{1}{n^{2}} as it tends to 0. = lim ⁡ n → ∞ n . s i n ( π n ) =\lim_{n\to\infty}n.sin(\dfrac{\pi}{n}) = π \boxed{=\pi}

26 Helpful 0 Interesting 1 Brilliant 0 Confused

I could not understand the last part that how it became equal to pi.

Naveen Kumar - 5 years, 5 months ago

Substituting 9999999999 as the value of infinity will yield the value of pi..

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice one!!

Aniswar S K - 4 years, 1 month ago

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