Originally Posted by

**Kataangel** I've got a question to *use the notion of continuity to show that *$\displaystyle x^7=x-2$* has at least one real root, that is, there is at least one value of x that solves this equation.*

If I try to zero one side, I'm still not sure how to get the roots"

$\displaystyle x^7-x-2=0$

I think it involves the continuity definition which I am aware of below (trying to get the x goes to a below lim)

$\displaystyle \lim{x \to a}f(x)=f(a)$

So I don't know how that will fit in with my problem above. How would I find the solution to it then?