- #1
kbrono
- 16
- 0
Show that the function f(x)=x is continuous at every point p.
Here's what I think but not sure if i can make one assumption.
Let [tex]\epsilon[/tex]>0 and let [tex]\delta[/tex]=[tex]\epsilon[/tex] such that for every x[tex]\in[/tex][tex]\Re[/tex] |x-p|<[tex]\delta[/tex]=[tex]\epsilon[/tex]. Now x=f(x) and p=f(p) so we have |f(x)-f(p)|<[tex]\epsilon[/tex].
Or...
can i just say that |x-p| [tex]\leq[/tex] |f(x)-f(p)|<[tex]\epsilon[/tex]. ?
Thanks
Here's what I think but not sure if i can make one assumption.
Let [tex]\epsilon[/tex]>0 and let [tex]\delta[/tex]=[tex]\epsilon[/tex] such that for every x[tex]\in[/tex][tex]\Re[/tex] |x-p|<[tex]\delta[/tex]=[tex]\epsilon[/tex]. Now x=f(x) and p=f(p) so we have |f(x)-f(p)|<[tex]\epsilon[/tex].
Or...
can i just say that |x-p| [tex]\leq[/tex] |f(x)-f(p)|<[tex]\epsilon[/tex]. ?
Thanks