Proposition:
Let (X,dx),(Y,dy),(Z,dz) be metric spaces. Let x0∈X,,y0∈Y,,z0∈Z. Let f: X→Y and g: Y→Z be functions,and let E be a set. If we have lim f(x) = y0 as x→x0 and x∈E and lim g(x) = z0 as y→y0 and y∈f(E). Conclude that lim gof(x) = z0 as x→x0 and x∈E.
This is my attempt.
Since lim f(x) = y0 as x→x0, for all ε1, there exists δ1 such that for all x∈E, whenever dx(x,x0)<δ1, we have dy(f(x),y0)<ε1.
Since lim g(x) = z0 as y→y0 and y∈f(E), for all ε2, there exists δ2=ε1 such that for all y∈f(E), whenever dy(f(x),y0)<δ2=ε1, we have dz(g(f(x)),z0)<ε2.
I think my proof is wrong in orders of ε and δ. How to prove this proposition? What's wrong with my proof? Thanks a lot!
Your puzzles are correct, and you basically miss how to put them together in a proper way.
As the formula to prove starts with '$\forall\varepsilon\,\exists\delta$', in the end, we want to find a $\delta>0$ for every $\varepsilon>0$.
Let $\varepsilon>0$ be given, then set $\varepsilon_2:=\varepsilon$ and find $\delta_2=:\varepsilon_1$ then $\delta_1=:\delta\ $ by 2. and 1.
Now, if $\ d(x,x_0)<\delta,\ $ then $\ d(f(x),y_0)<\varepsilon_1=\delta_2,\ $ so $\ d(g(f(x)),\,z_0)<\varepsilon_2=\varepsilon$.