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There is a specific sample question about limits that I haven't been able to solve.

The function $f$ can be written as $c^x/(c+n)^x$, where $c > 0, n>0$, so for example, something like $4^x/5^x$. The limit as $x$ tends to infinity is said to be $0$ for all of these functions, however this isn't simply obvious to me.

I can understand that the function will always be decreasing, and that even so, it cannot become negative, meaning that the limit as $x$ goes to infinity should be well defined.

I'm wondering then, if someone could show a proof that it does tend to $0$. Any thoughts/ideas would be really appreciated.

Sam
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3 Answers3

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Let $a=c<c+n=b$.

$$\frac{a^x}{b^x}=\left(\frac ab\right)^x$$

For $x\to\infty$, one might assume the limit exists and call it $L$. Then do some algebra:

$$L=\lim_{x\to\infty}\left(\frac ab\right)^x=\lim_{x\to\infty}\left(\frac ab\right)^{x+1}=\frac ab\lim_{x\to\infty}\left(\frac ab\right)^x=\frac abL$$

So if the limit existed, then

$$L=\frac abL$$

Which has the single real solution $L=0$.

We know that $L$ exists because, as you say, $\left(\frac ab\right)^x$ is decreasing in $x$ and bounded below by $0$, so it exists and

$$0\le L<\frac ab$$

Simply Beautiful Art
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    beautiful and simple answer. You honor your nick! :p –  Aug 05 '17 at 22:21
  • I don't get why we'd assume there is a limit - you can use Bernoulli's inequality and prove it directly. – Chris Aug 05 '17 at 22:22
  • @Chris ...Or you could read the last few lines for a much more elementary approach to showing the limit exists... – Simply Beautiful Art Aug 05 '17 at 22:23
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    @Chris right... but Bernoulli's inequality is not so elementary, and this limit is elementary. Probably it would hide some kind of circularity proving it through Bernoulli's path. –  Aug 05 '17 at 22:24
  • @SimplyBeautifulArt It's not really elementary to appeal to if the student doesn't follow it. The property you're alluding to (using the greatest lower bound) is probably more confusing for a beginning student than a simple, direct inequality. Although yes, it's nice in its own way. – Chris Aug 05 '17 at 22:29
  • @Masacroso there's nothing circular there - on the contrary, it's the standard proof. I picked it up from Rudin, actually. – Chris Aug 05 '17 at 22:31
  • @Masacroso: Bernoulli's inequality can be proven very simply by an easy induction proof. – robjohn Aug 05 '17 at 23:01
  • @robjohn Thanks for the link :) – Simply Beautiful Art Aug 05 '17 at 23:13
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Your fraction is $$e^{x \Bigl(\ln (c)-\ln (c+n)\Bigr)} $$

with $$\ln (c)-\ln (c+n)<0$$ since $x \mapsto \ln (x ) $ is strictly increasing at $\Bbb R^+$.

thus the limit is $e^{-\infty}=0$.

hamam_Abdallah
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In general what you're asking amounts to "Why does $a^x \to 0$ as $x \to \infty$ for $a<1$?" The simplest answer is that if $a < 1$ then $a = \frac{1}{1 + t}$ for some $t > 0$, and $(\frac{1}{1+t})^n < \frac{1}{1+nt}$, since $(1 +t)^n > 1 + nt$. For $x > n$, we therefore have $(\frac{1}{1+t})^x < \frac{1}{1 + nt}$, so as $x$ goes to infinity this is going to become arbitrarily small. In particular, if you pick $x > \frac{N}{t}$, then $a^x = (\frac{1}{1+t})^x < \frac{1}{1 + xt} < \frac{1}{N}$.

Chris
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