Behaviour of $x^n$ for very large $n$?

when I plotted the graphs of $y=x^{10},x^{100},x^{1000}$ etc. I noticed that the shape approached an open rectangle with base between $x=-1$ and $x=1$,
But why does $x^n$ approaches this shape and is almost zero for $x \in (-1,1)$ and increases suddenly afterwards.
Are there any other functions which change behaviour suddenly, Please explain...

• Because $x^n \to \infty$ for $|x| > 1$ and 0 for $|x| < 1$, as $n \to \infty$ Jun 6, 2018 at 8:19
• The reason why it is near zero for $x\in(-1,1)$ is that when you exponentiate within that radius , the larger the exponent the smaller the values get . if $-1\lt x\lt 1$ then for larger powers the values decrease. if $x = -1$ then the values alternate if $x= 1$ then it stays $1$ Jun 6, 2018 at 8:19
• You probably mean for $f_n\in C^0,f_n\to f$, $f\notin C^0$. You can also consider $f_n(x)=\frac1{1+x^n}$. Jun 6, 2018 at 8:21
• @LittleCuteKemono I saw the graph it looks like a box Jun 6, 2018 at 8:22
• @Mathstextbook Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– user
Aug 3, 2018 at 21:56

The function $x^n$ is the repeated multiplication of $x$ with itself. Let's focus on positive values $x>0$ first, the negatives are a little hassle but not too hard, if you understood the concept.

As long as $x<1$ holds, this value will get smaller every time you mutliply it. Take $0.9$ for example $$0.9>0.81>0.729>0.6561>...$$

The contrary is done, when $x>1$ holds true. In that case, you enlarge the value with each additional multiplication. Take $1.1$: $$1.1<1.21<1.331<1.4641<...$$ This difference will increase, as you increase $n$.

• Exactly correct (except we should really look at the magnitude, $|x|<1$). Just for some extra information for the OP, consider the boundary case where $x = 1$. No matter how many times you multiply $1$ with itself it remains $1$, i.e. $1\times 1\times 1 \times ...\times 1 = 1$. Therefore $x = 1$ is just the point that separates the two different behavioral patterns of the function.
– Eff
Jun 6, 2018 at 8:22
• Absolutly right, I will add a positivity-constraint to my answer. Jun 6, 2018 at 13:41

The fact is that

• for $|a|<1$ that is $-1<a<1$ we have $a^n\to 0$

whereas

• for $|a|>1$ that is $a<-1$ and $a>1$ we have $|a^n|\to \infty$