Hi, we are asked to prove that 0 raised to infinity is zero by using limit definitions such as:
limx→af(x)=0, limx→ag(x)=∞, and limx→af(x)^g(x)=∞ where f(x) is a positive function. Thanks in advance.
Since infinity is a symbol and not a number, there needs to be a definition for what you mean by taking 0 (or any other number) to the infinity power.
I think what you mean to say is that you want to prove that for any real number a, $\displaystyle -1<a<1$, $\displaystyle \lim_{n\to\infty} a^n = 0$. This should be pretty easy if you've done proofs about limits before. If not, you can just think about what happens if you keep multiplying over and over by a.
- Hollywood
Unfortunately though, we are only asked to prove that 0 raised to infinity is equal to zero and is not of indeterminate form. But we are given a hint to use these limit definitions:
\lim_{x \rightarrow a}f(x)=0, \lim_{x \rightarrow a}g(x)=\infty, and then evaluating \lim_{x \rightarrow a}f(x)^{g(x)}. I really wish it was more specific and yeah, it would be pretty easy if it was the same case as -1<a<1, \lim_{n \rightarrow \infty}a^n=0.
-Andoy
By the definition of $\displaystyle \lim_{x \rightarrow a}f(x)=0$, there is a $\displaystyle \delta_1>0$ such that whenever $\displaystyle |x-a|<\delta_1$, $\displaystyle |f(x)-f(0)|= |f(x)| < \frac{1}{2}$ (just set $\displaystyle \epsilon=\frac{1}{2}$ in the definition).
Now let $\displaystyle \epsilon>0$, and let M be such that $\displaystyle \frac{1}{2^M}<\epsilon$. By the definition of $\displaystyle \lim_{x \rightarrow a}g(x)=\infty$, there is a $\displaystyle \delta_2>0$ such that whenever $\displaystyle |x-a|<\delta_2$, $\displaystyle |g(x)|>M$. So let $\displaystyle \delta=\min(\delta_1,\delta_2)$. Then whenever $\displaystyle |x-a|<\delta$:
$\displaystyle |f(x)^{g(x)}|<\left|\left(\frac{1}{2}\right)^{g(x) }}\right|$ (since $\displaystyle |f(x)|<\frac{1}{2}$)
$\displaystyle <\left|\left(\frac{1}{2}\right)^M}\right|$ (since $\displaystyle |g(x)|>M$)
$\displaystyle <\epsilon$. Therefore $\displaystyle \lim_{x\to{a}}f(x)^{g(x)}=0$.
- Hollywood
Now there's where I think I screwed up before I posted this thread. I tried to prove it using the same method but without setting a constant for my Epsilon. I'm not really sure if I'm allowed to set my Epsilon to a constant but if I was, then you sure did help me lots. Thanks.
-Andoy
Hi Serena! That was a good idea, and by logic, we can say that 0 raised to infinity is technically 0 raised to a very large finite number since infinity is not a number. But unfortunately, the limit you provided just now do not prove anything that in a function k^n where k is approaching zero and n approaches infinity that the whole function itself would go to zero. Nice try though. What I did is I used the same proof of hollywood but I defined abs(f(x)-L)<Epsilon, as abs(f(x))<epsilon since L is defined zero in the limit as x approaches a. After that I used the squeeze theorem to basically say that f(x) approximates to (-E,E). From there on, I took the limit to infinity of approx f(x) as E because f(x) is assumed to be positive. From there I proved that Limit of f(x)^g(x) as x approaches infinity is zero. Thanks for reply guys, and feel free to say if I got anything wrong from my proof.
P.S. how do you guys write those neat equations right there? I tried latex but It doesn't seem to work. XD
Oh? I didn't see in your problem statement that you had to use a function k^n.
Seems like overcomplicating the problem.
If you click Go Advanced, type a latex formula, select it, and click the $\displaystyle \fbox{\sum}$ button, it'll become Latex.P.S. how do you guys write those neat equations right there? I tried latex but It doesn't seem to work. XD
Alternatively, just type [ TEX ] <your formula> [ /TEX ]. Without the spaces of course.
Or if you click Reply With Quote, you can see how someone else did it.