Calculation rules for limits (continuation)

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:

Calculate the following limit using the limit laws: $\lim_{n\to\infty} \frac{4 n + 3}{-2 n + 3}\text.$

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 $\frac{4 n + 3}{-2 n + 3} = \frac{4 + \dfrac{3}{n}}{-2 + \dfrac{3}{n}}\text.$ The limit rule for multiplication by constants gives that $\begin{aligned} \lim_{n\to\infty}\frac{3}{n} &= 3 \cdot \lim_{n\to\infty}\frac{1}{n} = 3 \cdot 0 =0\\[0.25cm] \lim_{n\to\infty}\frac{3}{n} &= 3 \cdot \lim_{n\to\infty}\frac{1}{n} = 3 \cdot 0 =0\text.\end{aligned}$ The sum rule for limits now gives $\begin{aligned} \lim_{n\to\infty} \left(4 + \frac{3}{n}\right) &= \lim_{n\to\infty} 4 + \lim_{n\to\infty}\frac{3}{n} = 4 \\[0.25cm] \lim_{n\to\infty} \left(-2 + \frac{3}{n}\right) &= \lim_{n\to\infty} -2 + \lim_{n\to\infty}\frac{3}{n} = -2\text.\end{aligned}$ The quotient rule for limits now gives that $\lim_{n\to\infty} \frac{\displaystyle 4 + \frac{3}{n}}{\displaystyle -2 + \frac{3}{n}} = \frac{4}{-2}\text.$ Therefore the answer is $-2$ .

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.