Collection of Maths Problems

Arithmetic of limits

Task number: 2866

• Variant 1

  \(\displaystyle \lim_{n\to\infty} \frac{n+1}{n+2}\)

• Hint

  Factor out \(n\).

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{n+1}{n+2} = \lim_{n\to\infty} \frac{n(1+\frac{1}{n})}{n(1+\frac{2}{n})} = \frac{\lim\limits_{n\to\infty} (1+\frac{1}{n})}{\lim\limits_{n\to\infty} (1+\frac{2}{n})} = \frac{\lim\limits_{n\to\infty} 1+ \lim\limits_{n\to\infty}(\frac{1}{n})}{\lim\limits_{n\to\infty} 1+2\lim\limits_{n\to\infty}(\frac{1}{n})} = \frac{1 + 0}{1 + 2{\cdot}0} = 1. \)

  We have used propositions about arithmetic limits in the steps above.

• Result

  The limit equals \(1\).

• Variant 2

  \(\displaystyle \lim_{n\to\infty} \left(4+\frac1n+\frac3{n^2-2n}\right)\left(5-\frac1{n^2}\right)\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \left(4+\frac1n+\frac3{n^2-2n}\right)\left(5-\frac1{n^2}\right)=\left(4+\lim_{n\to\infty} \frac1n+\lim_{n\to\infty} \frac3{n^2-2n}\right)\left(5-\lim_{n\to\infty} \frac1{n^2}\right) =\)

  \[\displaystyle =\left(4+0+\frac{\lim\limits_{n\to\infty} (\frac{3}{n^2})}{\lim\limits_{n\to\infty} (1-\frac{1}{n})}\right)\left(5-0\right)=\left(4+\frac{0}{1}\right)\cdot5=20\]

  We have used propositions about arithmetic limits in the steps above.

• Result

  The limit equals \(20\).

• Variant 3

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

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{3n^2+5n}{-n^2+4n} = \lim_{n\to\infty} \frac{n^2(3+\frac{5}{n})}{n^2(-1+\frac{4}{n})} = \frac{\lim\limits_{n\to\infty} (3+\frac{5}{n})}{\lim\limits_{n\to\infty} (-1+\frac{4}{n})} = \frac{3 + 0}{-1 + 0} = -3. \)

• Result

  The limit equals \(-3\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty} \frac{2n^2+n-3}{n^3-1}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{2n^2+n-3}{n^3-1} = \frac{\lim\limits_{n\to\infty} (\frac{2}{n}+\frac{1}{n^2}-\frac{3}{n^3})}{\lim\limits_{n\to\infty} (1-\frac{1}{n^3})} = \frac{0+0-0}{1-0}=0 \)

• Result

  The limit equals \(0\).

• Variant 5

  \(\displaystyle \lim_{n\to\infty} \frac{2n^3+6n}{n^3-7n+7}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{2n^3+6n}{n^3-7n+7} = \frac{\lim\limits_{n\to\infty} (2+\frac{6}{n^2})}{\lim\limits_{n\to\infty} (1-\frac{7}{n^2}+\frac{7}{n^3})} = \frac{2+0}{1-0+0}=2 \)

• Result

  The limit equals \(2\).

• Variant 6

  \(\displaystyle \lim_{n\to\infty} \frac{n^3-1}{2n^2+n-3}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{n^3-1}{2n^2+n-3}= \frac{\lim\limits\limits_{n\to\infty} (n-\frac{1}{n^2})}{\lim\limits_{n\to\infty} (2+\frac{1}{n}-\frac{3}{n^2})}= \frac{\infty-0}{2+0-0}=\infty \)

• Result

  The sequence has the improper limit \(\infty\).

• Variant 7

  \(\displaystyle \lim_{n\to\infty} \frac{\sqrt n}{n^3+1}\)

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{\sqrt n}{n^3+1}= \frac{\lim\limits_{n\to\infty} n^{-\frac{5}{2}}}{\lim\limits_{n\to\infty} (1+\frac{1}{n^3})}= \frac{0}{1+0}=0 \)

• Result

  The limit equals \(0\).

• Variant 8

  \(\displaystyle \lim_{n\to\infty} \frac{(2n-3)^{20}(3n+2)^{30}}{(2n+1)^{50}}\)

• Hint

  Given the rule for the limit of a sum, a rule for the limit of a power follows.

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{(2n-3)^{20}(3n+2)^{30}}{(2n+1)^{50}}= \lim_{n\to\infty} \frac{n^{20}(2-\frac{3}{n})^{20}n^{30}(3+\frac{2}{n})^{30}}{n^{50}(2+\frac{1}{n})^{50}}=\)

  \[=\frac{\lim\limits_{n\to\infty}(2-\frac{3}{n})^{20} \lim\limits_{n\to\infty} (3+\frac{2}{n})^{30}}{\lim\limits_{n\to\infty}(2+\frac{1}{n})^{50}}= \frac{2^{20} 3^{30}}{2^{50}}=\left(\frac32\right)^{30}\]

• Result

  The limit equals \(\displaystyle\left(\frac32\right)^{30}\).

• Variant 9

  \(\displaystyle \lim_{n\to\infty} \frac{3^n+5^n+10^n}{-2^{n+1}+5^{n+1}+10^{n+1}}\)

• Hint

  Factor out \(\displaystyle10^n\).

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{3^n+5^n+10^n}{-2^{n+1}+5^{n+1}+10^{n+1}} = \lim_{n\to\infty} \frac{10^n((\frac{3}{10})^n+(\frac{5}{10})^n+1)}{10^n(-2(\frac{2}{10})^n+5(\frac{5}{10})^n+10)} = \frac{0 + 0 + 1}{-2{\cdot} 0 + 5{\cdot} 0 + 10} = \frac1{10}. \)

  We have simplified the last step a bit, but you should be aware that we have used propositions about arithmetic limits in it.

• Result

  The limit equals \(\displaystyle\frac1{10}\).