- #1
MIB
- 17
- 0
This theorem I understand but I was only stuck in a simple implication in the proof
Theorem : If f is continuous at b and [itex]{lim}_{x \rightarrow a} g(x) = b [/itex] ,
then , [itex]{lim}_{x \rightarrow a} f(g(x)) = f(b)[/itex] .
proof
since f is continuous at b then , Given [itex]\epsilon > 0 [/itex] , there exists [itex]\delta_1 > 0[/itex] such that ,
if [itex]0 < |y-b|<\delta[/itex] then [itex]|f(x)-f(b)|<\epsilon[/itex] ... 1
and since [itex]{lim}_{x \rightarrow a} g(x) = b [/itex] , then ther exist δ such that ,
if [itex]0 < |x-a|<\delta[/itex] then [itex]|g(x)-b|<\delta_1[/itex]
it is easy to show that f is defined on some open interval containing a , the proplem is in the following implication ,
that is [itex]|g(x)-b|<\delta_1[/itex] implies [itex]|f(x)-f(b)|<\epsilon[/itex]
Here I see that this true only if g(x) is in the domain of f to replace it with y that is there is some y such that y=g(x) where g(x) must takes all values in some interval containing b ( Right) , I see tat is true because because we can make [itex]|g(x)-b|<\epsilon[/itex] for arbitrary ε , that is even for very small ε so we must have then g(x) takes all values on some interval containing b . (Right)
Thanks
sorry I posted before I finished
Mod note: For absolute values, just use two | characters.
Theorem : If f is continuous at b and [itex]{lim}_{x \rightarrow a} g(x) = b [/itex] ,
then , [itex]{lim}_{x \rightarrow a} f(g(x)) = f(b)[/itex] .
proof
since f is continuous at b then , Given [itex]\epsilon > 0 [/itex] , there exists [itex]\delta_1 > 0[/itex] such that ,
if [itex]0 < |y-b|<\delta[/itex] then [itex]|f(x)-f(b)|<\epsilon[/itex] ... 1
and since [itex]{lim}_{x \rightarrow a} g(x) = b [/itex] , then ther exist δ such that ,
if [itex]0 < |x-a|<\delta[/itex] then [itex]|g(x)-b|<\delta_1[/itex]
it is easy to show that f is defined on some open interval containing a , the proplem is in the following implication ,
that is [itex]|g(x)-b|<\delta_1[/itex] implies [itex]|f(x)-f(b)|<\epsilon[/itex]
Here I see that this true only if g(x) is in the domain of f to replace it with y that is there is some y such that y=g(x) where g(x) must takes all values in some interval containing b ( Right) , I see tat is true because because we can make [itex]|g(x)-b|<\epsilon[/itex] for arbitrary ε , that is even for very small ε so we must have then g(x) takes all values on some interval containing b . (Right)
Thanks
sorry I posted before I finished
Mod note: For absolute values, just use two | characters.
Last edited by a moderator: