It must not exist.

Calculus Level 2

When n n is a positive integer, what is the value of

lim ⁡ n → ∞ ( − 1 ) n − 1 sin ⁡ ( π n 2 + 0.5 n + 1 ) ? \lim_{n \rightarrow \infty} ( -1 ) ^{n-1} \sin ( \pi \sqrt{ n^2 + 0.5 n + 1 } ) ?

sin ⁡ π 3 \sin\frac{\pi}{3} − 1 2 -\frac{1}{\sqrt{2}} 0 0 The limit does not exist.

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Sanjeet Raria by Sanjeet Raria , 27, India

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2 solutions

U Z
Jan 29, 2015

lim ⁡ n → ∞ − ( − 1 ) n ( − 1 ) n − 1 sin ⁡ ( n π − π n 2 + 0.5 n + 1 ) \lim_{n \rightarrow \infty} - ( -1 ) ^{n} (-1)^{n - 1} \sin \left( n\pi - \pi \sqrt{ n^2 + 0.5 n + 1 } \right)

lim ⁡ n → ∞ − ( − 1 ) 2 n − 1 sin ⁡ π ( ( n − n 2 + n 2 + 1 ) ( n + n 2 + n 2 + 1 ) n + n 2 + n 2 + 1 ) \lim_{n \rightarrow \infty} - (-1)^{2n - 1} \sin\pi \left( \dfrac{\left( n - \sqrt{ n^2 + \dfrac{n}{2} + 1 } \right) \left( n + \sqrt{ n^2 + \dfrac{n}{2} + 1 } \right)}{ n + \sqrt{ n^2 + \dfrac{n}{2} + 1 }}\right)

= lim ⁡ n → ∞ − ( − 1 ) 2 n − 1 sin ⁡ π ( n 2 − n 2 − n 2 − 1 n ( 1 + 1 + 1 2 n + 1 n 2 ) ) = \lim_{n \rightarrow \infty} - (-1)^{2n - 1} \sin\pi \left( \dfrac{n^2 - n^2 - \dfrac{n}{2} - 1}{n \left(1 + \sqrt{1 + \dfrac{1}{2n} + \dfrac{1}{n^2}}\right)}\right)

= lim ⁡ n → ∞ − ( − 1 ) 2 n sin ⁡ π ( 1 2 + 1 n 1 + 1 + 1 2 n + 1 n 2 ) = \lim_{n \rightarrow \infty} - (-1)^{2n} \sin\pi\left(\dfrac{\dfrac{1}{2} + \dfrac{1}{n}}{1 + \sqrt{1 + \dfrac{1}{2n} + \dfrac{1}{n^2}}}\right)

= − 1 × sin ⁡ π 4 = − 1 2 = -1 \times \sin\dfrac{\pi}{4} = - \dfrac{1}{\sqrt{2}}

22 Helpful 3 Interesting 0 Brilliant 1 Confused

Hi , in the very first line of your solution how did you write − 1 × ( − 1 ) n -1 \times (-1)^{n} since you are introducing a -ve sign . Also ( − 1 ) n (-1)^{n} can't be -ve since it is not given if n is odd so even it can't be -ve

Kudou Shinichi - 6 years, 4 months ago

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see the question we are given ( − 1 ) n − 1 \dfrac{(-1)^n}{-1}

U Z - 6 years, 4 months ago

Nice solution

Dunstan lame - 6 years, 3 months ago

What's wrong with this:

First we note that sin ⁡ ( π ( n 2 + 0.5 n + 1 ) 1 / 2 ) = sin ⁡ ( π n ( 1 + ( 2 n ) − 1 + n − 2 ) 1 / 2 ) \sin({\pi (n^2+0.5n+1)^{1/2} })= \sin({\pi n (1+(2n)^{-1}+n^{-2} )^{1/2}})

As a second observation ∣ ( − 1 ) n − 1 sin ⁡ ( π n ( 1 + ( 2 n ) − 1 + n − 2 ) ) ∣ = ∣ sin ⁡ ( π n ( 1 + ( 2 n ) − 1 + n − 2 ) ) ∣ ≤ 1. |(-1)^{n-1} \sin({\pi n (1+(2n)^{-1}+n^{-2} )} )| = |\sin({\pi n (1+(2n)^{-1}+n^{-2} )} )| \leq 1.

With this two observations in hand lim ⁡ n → ∞ ∣ sin ⁡ ( π n ( 1 + ( 2 n ) − 1 + n − 2 ) ) ∣ = ∣ lim ⁡ n → ∞ sin ⁡ ( π n ( 1 + ( 2 n ) − 1 + n − 2 ) ) ∣ Continuity = ∣ sin ⁡ ( lim ⁡ n → ∞ π n ( 1 + ( 2 n ) − 1 + n − 2 ) ) ∣ Continuity = ∣ sin ⁡ ( lim ⁡ n → ∞ π n × lim ⁡ n → ∞ ( 1 + ( 2 n ) − 1 + n − 2 ) ) ∣ = ∣ sin ⁡ ( lim ⁡ n → ∞ π n ) ∣ = ∣ lim ⁡ n → ∞ sin ⁡ ( π n ) ∣ Continuity = 0 n is integer. \begin{aligned} \lim_{n\rightarrow \infty} |\sin ({\pi n (1+(2n)^{-1}+n^{-2} )}) | &=|\lim_{n\rightarrow \infty} \sin({\pi n (1+(2n)^{-1}+n^{-2} )}) | \qquad \text{Continuity}\\ & =| \sin({\lim_{n\rightarrow \infty} \pi n (1+(2n)^{-1}+n^{-2} )} )|\qquad \text{Continuity}\\ &=| \sin({\lim_{n\rightarrow \infty} \pi n \times \lim_{n\rightarrow \infty} (1+(2n)^{-1}+n^{-2} )}) |\\ & =| \sin ({\lim_{n\rightarrow \infty} \pi n} )| \\ &=| \lim_{n\rightarrow \infty} \sin({\pi n} )| \qquad \text{Continuity}\\ &=0 \qquad n \text{ is integer.} \end{aligned}

Gerónimo Rojas Barragán - 4 years, 10 months ago

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@Gerónimo Rojas Barragán

Your mistake is in lim ⁡ n → ∞ n ( 1 + ( 2 n ) − 1 + ⋯ ) = lim ⁡ n → ∞ n × lim ⁡ n → ∞ ( 1 + ( 2 n ) − 1 + ⋯ ) \lim_{n\to\infty}n(1+(2n)^{-1}+\cdots) = \lim_{n\to\infty}n\times\lim{n\to\infty}(1+(2n)^{-1}+\cdots) .

Also, you lost a factor 1 2 \frac{1}{2} along the way when getting rid of the square root. It should become ( 4 n ) − 1 (4n)^{-1} .

Tom Verhoeff - 4 years, 3 months ago

What was your motivation in writing sin in npi - theta form

Ameya Anjarlekar - 4 years, 9 months ago
Tom Verhoeff
Mar 2, 2017

A more concise solution (using that n n is a positive integer): ( − 1 ) n − 1 sin ⁡ ( π n 2 + 0.5 n + 1 ) = (-1)^{n-1}\sin\big(\pi\sqrt{n^2 + 0.5n + 1}\big) = ( − 1 ) n − 1 sin ⁡ ( π n 1 + 0.5 / n + 1 / n 2 ) = (-1)^{n-1}\sin\big(\pi n\sqrt{1 + 0.5/n + 1/n^2}\big) = ( − 1 ) n − 1 sin ⁡ ( π n ( 1 + 1 2 0.5 / n + O ( 1 / n 2 ) ) = (-1)^{n-1}\sin\big(\pi n(1 + \frac{1}{2}0.5/n + \mathcal{O}(1/n^2)\big) = ( − 1 ) n − 1 sin ⁡ ( π n + π / 4 + O ( 1 / n ) ) = (-1)^{n-1}\sin\big(\pi n + \pi/4 + \mathcal{O}(1/n)\big) = − sin ⁡ ( π / 4 + O ( 1 / n ) ) → -\sin\big(\pi/4 + \mathcal{O}(1/n)\big) \to − sin ⁡ ( π / 4 ) = -\sin(\pi/4) = − 1 2 -\frac{1}{\sqrt{2}}

8 Helpful 0 Interesting 2 Brilliant 2 Confused

What’s wrong with saying that sin ⁡ ( π n 2 + 0.5 n + 1 ) → sin ⁡ ( π n 2 ) = sin ⁡ ( n π ) = 0 \sin\left(\pi\sqrt{n^2 +0.5n+1 } \right) \to \sin\left(\pi\sqrt{n^2}\right) = \sin(n\pi) = 0 ?

Tavish Music - 6 months, 2 weeks ago

yes exctly the same question as above, why cant I say, lim ⁡ n → ∞ sin ⁡ ( π n 1 + 0.5 n + 1 n 2 ) = 0 \lim_{n\rightarrow\infty}\displaystyle\sin\left(\pi n\sqrt{1+\frac{0.5}{n}+\frac{1}{n^2}}\right)=0

Sarthak Sahoo - 1 month, 3 weeks ago

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