18. Sequences

b5e. Indeterminate Forms \(0^0\) and \(1^\infty\) and \(\infty^0\)

Recall that \(e^x\) and \(\ln(x)\) are inverse functions. Further, recall that limits can be moved inside continuous functions.

Method:

1. These sequences are all of the form: \((a_n)^{b_n}\). We apply \(e^{\ln{}}\) to each term, move the limit inside the exponential function (since it is continuous) and pull the exponent out of the log (using the identity \(\ln{a^b}=b\ln{a}\)). This will convert the limit to the indeterminate form \(0\cdot\infty\).

The first example has indeterminate form \(1^\infty\).

Compute \(\lim\limits_{n\to\infty} \left(1+\dfrac{2}{n}\right)^{3n}\).

We first apply \(e^{\ln{}}\) to the term: \[ \lim_{n\to\infty} \left(1+\dfrac{2}{n}\right)^{3n} =\lim_{n\to\infty} {\large e}^{\displaystyle \ln{\left(1+\dfrac{2}{n}\right)^{3n}}} \] Next we move the limit inside the exponential (since it is a continuous function) and pull the exponent out the log: \[ \lim_{n\to\infty} \left(1+\dfrac{2}{n}\right)^{3n} ={\large e}^{\displaystyle \lim_{n\to\infty} 3n\ln{\left(1+\dfrac{2}{n}\right)}} \] The new limit has the form \(0\cdot\infty\). We convert it to \(\dfrac{0}{0}\) by writing the \(n\) factor as \(\dfrac{1}{n}\) in the denominator. Then we apply l'Hopital's Rule and simplify: \[\begin{aligned} \lim_{n\to\infty} \left(1+\dfrac{2}{n}\right)^{3n} &={\large e}^{\displaystyle \lim_{n\to\infty} \dfrac{3\ln{\left(1+\dfrac{2}{n}\right)}}{\dfrac{1}{n}}} \\ &\overset{\text{l'H}}{=}{\large e}^{\displaystyle \lim_{n\to\infty} \dfrac{\dfrac{3}{1+\dfrac{2}{n}}\left(\dfrac{-2}{n^2}\right)}{-\dfrac{1}{n^2}}} \\ &={\large e}^{\displaystyle \lim_{n\to\infty} \dfrac{6}{1+\dfrac{2}{n}}} \\[5pt] &={\large e^6} \end{aligned}\]

This exercise has indeterminate form \(\infty^0\).

This exercise has indeterminate form \(0^0\).