Collection of Maths Problems

Limits of sequences

Task number: 2870

• Variant 1

  \(\displaystyle \lim_{n\to\infty} \frac{2n^2+4n+n \sin n}{ n \cos 3n + (2n +\sin n)^2}\)

• Resolution

  The \(\sin\) and \(\cos\) functions are bounded, so \(\displaystyle \lim_{n\to\infty} \frac{\sin n}n = \lim_{n\to\infty} \frac{\cos 3n}n = 0\).

  \(\displaystyle \lim_{n\to\infty} \frac{2n^2+4n+n \sin n}{ n \cos 3n + (2n +\sin n)^2} = \lim_{n\to\infty} \frac{2 +4 \frac{1}{n} + \frac{\sin n}{n}}{\frac{\cos 3n}{n} + 4 + 4 \frac{\sin n}{n} + (\frac{\sin n}{n})^2} = \frac{2 +4 {\cdot} 0 + 0}{0 + 4 + 4 {\cdot} 0 + 0^2} =\frac12 \)

• Result

  The limit equals \(\displaystyle\frac12\).

• Variant 2

  \(\displaystyle \lim_{n\to\infty} \frac{\lfloor \sqrt{n} \rfloor}{\sqrt{n}}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{\lfloor \sqrt{n} \rfloor}{\sqrt{n}} \le \lim_{n\to\infty} \frac{\sqrt{n}}{\sqrt{n}} =1 \) and also \(\displaystyle \lim_{n\to\infty} \frac{\lfloor \sqrt{n} \rfloor}{\sqrt{n}} \ge \lim_{n\to\infty} \frac{\sqrt{n}-1}{\sqrt{n}}= \lim_{n\to\infty} (1-n^{-1/2})= 1 \).

• Result

  The limit equals \(1\).

• Variant 3

  \(\displaystyle \lim_{n\to\infty} \cos (n^2 \pi) +\cos ((n+1) \pi) \)

• Hint

  For odd \(n\) we have \(\cos (n^2 \pi) +\cos ((n+1) \pi) =-1+1=0\).

  For even \(n\) we have \(\cos (n^2 \pi) +\cos ((n+1) \pi) =1-1=0\).

  This is a constant sequence.

• Result

  The limit equals \(0\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty} n^{\cos (n \pi)} \)

• Resolution

  For even \(n\) we have \(n^{\cos (n \pi)}=n\), for odd \(n^{\cos (n\pi)}=\frac1n\).

  The first subsequence converges to \(\infty\), the second to 0.

• Result

  The sequence diverges; the limit does not exist.

• Variant 5

  \(\displaystyle \lim_{n\to\infty} \frac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt{n^3+1}}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt{n^3+1}} = \lim_{n\to\infty} \frac{n^\frac{2}{3}\sqrt[4]{1+\frac{2}{n^5}}-n^{\frac{2}{3}}\sqrt[3]{1+\frac{1}{n^2}}}{n^{\frac{4}{5}}\sqrt[5]{1+\frac{2}{n^4}}-n^{\frac{3}{2}}\sqrt{1+\frac{1}{n^3}}} =\)

  \[ =\lim_{n\to\infty} \frac{n^{\frac{5}{4}}\left(\sqrt[4]{1+\frac{2}{n^5}}-n^{-\frac{7}{12}}\sqrt[3]{1+\frac{1}{n^2}}\right)}{n^{\frac{3}{2}}\left(n^{-\frac{7}{10}}\sqrt[5]{1+\frac{2}{n^4}}-\sqrt{1+\frac{1}{n^3}}\right)} = \lim_{n\to\infty} n^{-\frac{1}{4}} \frac{\sqrt[4]{1+\frac{2}{n^5}}-n^{-\frac{7}{12}}\sqrt[3]{1+\frac{1}{n^2}}}{n^{-\frac{7}{10}}\sqrt[5]{1+\frac{2}{n^4}}-\sqrt{1+\frac{1}{n^3}}} =\] \[ =0 \cdot \frac{\sqrt[4]{1+0}-0\cdot \sqrt[3]{1+0}}{0\cdot \sqrt[5]{1+0}-\sqrt{1+0}} = 0 \]

• Result

  The limit equals \(0\).

• Variant 6

  \(\displaystyle \lim_{n\to\infty} \frac{3^n+n^5}{n^6+n!}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{3^n+n^5}{n^6+n!}= \lim_{n\to\infty} \frac{\frac{3^n}{n!}+\frac{n^5}{n!}}{\frac{n^6}{n!}+1}= \frac{0+0}{0+1}=0 \)

• Result

  The limit equals \(0\).

• Variant 7

  \(\displaystyle \lim_{n\to\infty} \frac{\sqrt[3]{n^2}\sin(n!)}{n+1}\)

• Resolution

  \(\displaystyle \left|\lim_{n\to\infty} \frac{\sqrt[3]{n^2}\sin(n!)}{n+1}\right|= \lim_{n\to\infty} \frac{\frac1{\sqrt[3]{n}}}{1+\frac1n}\left|\sin(n!)\right|\le \frac{0}{1+0}\cdot 1=0\).

• Result

  The limit equals \(0\).