Limits part 1: Infinite sequences: Basic calculation rules for limits

Theory 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:

Calculate the following limit using the limit laws: \[ \lim_{n\to\infty} \frac{-4 n -1}{3 n + 5}\text. \]
This is the indeterminate form \(\frac{\infty}{\infty}\). Now we have to rewrite this in order to obtain an expression in which we can apply the rules for limits.
If we divide the numerator and denominator by \(n\) we see that \[ \frac{-4 n -1}{3 n + 5} = \frac{-4 + \dfrac{-1}{n}}{3 + \dfrac{5}{n}}\text. \] The limit rule for multiplication by constants gives that \[ \begin{aligned} \lim_{n\to\infty}\frac{-1}{n} &= -1 \cdot \lim_{n\to\infty}\frac{1}{n} = -1 \cdot 0 =0\\[0.25cm] \lim_{n\to\infty}\frac{5}{n} &= 5 \cdot \lim_{n\to\infty}\frac{1}{n} = 5 \cdot 0 =0\text.\end{aligned} \] The sum rule for limits now gives \[ \begin{aligned} \lim_{n\to\infty} \left(-4 + \frac{-1}{n}\right) &= \lim_{n\to\infty} -4 + \lim_{n\to\infty}\frac{-1}{n} = -4 \\[0.25cm] \lim_{n\to\infty} \left(3 + \frac{5}{n}\right) &= \lim_{n\to\infty} 3 + \lim_{n\to\infty}\frac{5}{n} = 3\text.\end{aligned} \] The quotient rule for limits now gives that \[ \lim_{n\to\infty} \frac{\displaystyle -4 + \frac{-1}{n}}{\displaystyle 3 + \frac{5}{n}} = \frac{-4}{3}\text.\] Therefore the answer is \(-\frac{4}{3}\).
New example

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.

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