Collection of Maths Problems

Limits of n-th powers

Task number: 2873

• Variant 1

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

• Resolution

  \(\displaystyle \lim_{n\to\infty} \left(1+\frac{2}n\right)^n = \lim_{n\to\infty} \left(\left(1+\frac{1}{\frac12n}\right)^{\frac12n}\right)^2 = e^2 \)

• Result

  The limit equals \(e^2\).

• Variant 2

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

• Resolution

  \(\displaystyle \lim_{n\to\infty} \left(1+\frac{1}n\right)^{3n} = \lim_{n\to\infty} \left(\left(1+\frac{1}n\right)^{n}\right)^{3} =e^3 \)

• Result

  The limit equals \(e^3\).

• Variant 3

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

• Resolution

  \(\displaystyle \lim_{n\to\infty} \left(1+\frac{1}{n^2}\right)^n = \lim_{n\to\infty} \sqrt[n]{\left(1+\frac{1}{n^2}\right)^{n^2}}= \lim_{n\to\infty} \sqrt[n]{e}=1 \)

• Result

  The limit equals \(1\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty} \left(1-\frac{3}n\right)^n\)

• Hint

  Use a result from the preceding variations.

• Resolution

  \(\displaystyle \lim_{n\to\infty} \left(1-\frac{3}n\right)^n = \lim_{n\to\infty} \frac{\left(1-\frac{9}{n^2}\right)^n}{\left(1+\frac{3}n\right)^n} = \frac{1}{\lim\limits_{n\to\infty} \left(1+\frac{3}n\right)^n} = e^{-3} \)

  Alternately we may use the substitution \(k=n-3\)

  \(\displaystyle \lim_{n\to\infty} \left(1-\frac{3}n\right)^n = \lim_{n\to\infty} \left(\frac{n-3}n\right)^n = \lim_{k\to\infty} \left(\frac{k}{k+3}\right)^{k+3} = \frac1{ \lim\limits_{k\to\infty} \left(1+\frac3{k}\right)^{k}} \cdot \lim_{k\to\infty} \left(\frac{k}{k+3}\right)^{3} = \frac1{e^{3}} \cdot 1 \)

• Result

  The limit equals \(e^{-3}\).

• Variant 5

  \(\displaystyle \lim_{n\to\infty} \left(\frac{n+2}{n+5}\right)^{n+7}\)

• Resolution

  Using the approach from the preceding variant,

  \(\displaystyle \lim_{n\to\infty} \left(\frac{n+2}{n+5}\right)^{n+7}= \lim_{n\to\infty} \left(1-\frac{3}{n+5}\right)^{n+5}\left(\frac{n+2}{n+5}\right)^{2}=e^{-3}\cdot 1 \)

  Alternatively, dividing the numerator and denominator by \(n\):

  \(\displaystyle \lim_{n\to\infty} \left(\frac{n+2}{n+5}\right)^{n+7}= \frac {\lim\limits_{n\to\infty} \left(\frac{n+2}{n}\right)^{n+7}} {\lim\limits_{n\to\infty} \left(\frac{n+5}{n}\right)^{n+7}}= \frac{e^2}{e^5}=e^{-3} \)

• Result

  The limit equals \(e^{-3}\).