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...

  • $\begingroup$ Because $x^n \to \infty$ for $|x| > 1$ and 0 for $|x| < 1$, as $n \to \infty$ $\endgroup$
    – JuliusL33t
    Jun 6, 2018 at 8:19
  • $\begingroup$ 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 $1$ $\endgroup$ Jun 6, 2018 at 8:19
  • $\begingroup$ You probably mean for $f_n\in C^0,f_n\to f$, $f\notin C^0$. You can also consider $f_n(x)=\frac1{1+x^n}$. $\endgroup$ Jun 6, 2018 at 8:21
  • $\begingroup$ @LittleCuteKemono I saw the graph it looks like a box $\endgroup$ Jun 6, 2018 at 8:22
  • $\begingroup$ @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/… $\endgroup$
    – user
    Aug 3, 2018 at 21:56

2 Answers 2


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 $n$.

  • $\begingroup$ 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. $\endgroup$
    – Eff
    Jun 6, 2018 at 8:22
  • $\begingroup$ Absolutly right, I will add a positivity-constraint to my answer. $\endgroup$
    – Laray
    Jun 6, 2018 at 13:41

The fact is that

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


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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.