I'm trying to solve the limit of the following indeterminate form:

$$\lim_{t\to\infty} \frac{a+b(-m)^t}{c+d(-m)^{t-1}}$$

where $t=1, 2, 3, \cdots$ denotes time and all the coefficients are positive rational numbers.

In particular, I would like to consider two cases: i) $0<m<1$ and ii) $1<m$.

Any comments would be greatly appreciated!

  • $\begingroup$ Where you got this problem ? $\endgroup$ – Aakash Kumar Jul 24 '16 at 6:10
  • $\begingroup$ @AakashKumar: It is a model I built in my own work. $\endgroup$ – ppp Jul 24 '16 at 6:17

Hint :

i) $0\lt m\lt 1$

Since $-1\lt -m\lt 0$, we have $(-m)^{t}\to 0$ as $t\to \infty$.

ii) $1\lt m$

Since $-1\lt\frac{1}{-m}\lt 0$, we have $\frac{1}{(-m)^{t-1}}\to 0$ as $t\to\infty$. Now, divide both the numerator and the denominator by $(-m)^{t-1}$.

  • $\begingroup$ I have made a slight revision and some clarification. Can you please help me once again? Thanks much in advance! $\endgroup$ – ppp Jul 24 '16 at 11:51
  • $\begingroup$ @ppp: Ah, I didn't notice the issue. See the addition in my answer. $\endgroup$ – mathlove Jul 24 '16 at 11:56
  • $\begingroup$ Thanks! You used l'hopital's rule, right? But according to the definition of l'hopital's rule, it cannot be applied in my case as both numerator and denominator have to converge to either zero or infinity, no? $\endgroup$ – ppp Jul 24 '16 at 11:59
  • $\begingroup$ @ppp: No, we don't need the rule. For the first, the limit is $\frac{a+0}{c+0}=\frac ac$. For the second, the limit is $\frac{0+b(-m)}{0+d}=-\frac{bm}{d}$ $\endgroup$ – mathlove Jul 24 '16 at 12:01
  • $\begingroup$ So the point is that the limit operator can be separately applied to each relevant term in the numerator and denominator, right? $\endgroup$ – ppp Jul 24 '16 at 12:03

if $m>0$ we have $-m<0$ and if $x \in \mathbb{R}$ what kind of number is $$(-m)^{x}$$? in this form makes the limit no sense.

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Najib Idrissi Jul 24 '16 at 7:22
  • $\begingroup$ why not? $(-m)^x$ is not defined for $-m<0$ and $x\in \mathbb{R}$!! $\endgroup$ – Dr. Sonnhard Graubner Jul 24 '16 at 7:24
  • $\begingroup$ for example calculate $(-3)^{\sqrt{2}}$ $\endgroup$ – Dr. Sonnhard Graubner Jul 24 '16 at 7:25
  • $\begingroup$ Yes, I know it's not defined. But when a question doesn't make sense you comment on it, you don't try to answer it. $\endgroup$ – Najib Idrissi Jul 24 '16 at 7:26
  • $\begingroup$ Edit made to replace $x$ with $t$, discrete time index and to clarify that all the coefficients are positive rational numbers. $\endgroup$ – ppp Jul 24 '16 at 7:45

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