Limits part 1: Infinite sequences: Basic calculation rules for limits
Calculation rules for limits (continuation)
We have seen the following rules for calculating limits:
\[ \begin{aligned} \lim_{n\to\infty}\left( a_n + b_n\right) &= \lim_{n\to\infty} a_n + \lim_{n\to\infty} b_n \\[0.25cm] \lim_{n\to\infty}\left( a_n \cdot b_n\right) &= \lim_{n\to\infty} a_n \cdot \lim_{n\to\infty} b_n\\[0.25cm]\lim_{n\to\infty}(c\cdot a_n) &=c \cdot \lim_{n\to\infty} a_n\\[0.25cm] \lim_{n\to\infty} \frac{a_n}{b_n} &= \frac{\displaystyle\lim_{n\to\infty} a_n}{\displaystyle \lim_{n\to\infty} b_n}\end{aligned}\]
What happens with these rules for limits when these sequences diverge to plus of minus infinity? The answer is that the rules remain correct with the following conventions: \[\begin{aligned}
\infty + k &\ ``="\ \infty\\[0.25cm]
\infty + \infty &\ ``="\ \infty \\[0.25cm]
k\cdot\infty &\ ``="\ \infty \text{ als } k> 0\\[0.25cm]
k\cdot\infty &\ ``="\ -\infty \text{ als } k < 0\\[0.25cm]
\frac{k}{\infty} &\ ``="\ 0
\end{aligned}\] and similarly for \(-\infty\). If we would take for instance \(a_n = n\) and \(k = 2\) then \[ \lim_{n\to\infty} 2n = 2 \cdot\infty = \infty\text.\]
Note that we say nothing about expressions like \[ 0 \cdot\infty, \infty - \infty\text{ and } \frac{\infty}{\infty}\text.\] These expressions are called indeterminate forms. Even if we know that \(\lim_{n\to\infty} a_n = \infty\) and \(\lim_{n\to\infty} b_n = \infty\) we cannot say anything about \(\frac{a_n}{b_n}\). In fact, we already saw an exercise about this:
If we divide the numerator and denominator by \(n\) we see that \[ \frac{-4 n -4}{3 n -2} = \frac{-4 + \dfrac{-4}{n}}{3 + \dfrac{-2}{n}}\text. \] The limit rule for multiplication by constants gives that \[ \begin{aligned} \lim_{n\to\infty}\frac{-4}{n} &= -4 \cdot \lim_{n\to\infty}\frac{1}{n} = -4 \cdot 0 =0\\[0.25cm] \lim_{n\to\infty}\frac{-2}{n} &= -2 \cdot \lim_{n\to\infty}\frac{1}{n} = -2 \cdot 0 =0\text.\end{aligned} \] The sum rule for limits now gives \[ \begin{aligned} \lim_{n\to\infty} \left(-4 + \frac{-4}{n}\right) &= \lim_{n\to\infty} -4 + \lim_{n\to\infty}\frac{-4}{n} = -4 \\[0.25cm] \lim_{n\to\infty} \left(3 + \frac{-2}{n}\right) &= \lim_{n\to\infty} 3 + \lim_{n\to\infty}\frac{-2}{n} = 3\text.\end{aligned} \] The quotient rule for limits now gives that \[ \lim_{n\to\infty} \frac{\displaystyle -4 + \frac{-4}{n}}{\displaystyle 3 + \frac{-2}{n}} = \frac{-4}{3}\text.\] Therefore the answer is \(-\frac{4}{3}\).
Most indeterminate forms can be rewritten to different expressions in such a way that it is possible to apply the limit laws. When trying to determine a limit of the form \(\frac{a_n}{b_n}\) it helps sometimes to multiply the numerator and denominator by the same term so that the numerator or denominator converges to a finite number. Often it is a good idea to divide the numerator and denominator by the highest power of \(n\) in the denominator.
