Jtdylktuy

So I'm having a problem trying to make sense of a small detail in proving the following theorem:

To prove that if $displaystylelim_{xto a}f(x)=L$ and $displaystylelim_{xto b}g(x)=a$, then $displaystylelim_{xto b}f(g(x))=lim_{yto a}f(y)=L$, we need to find a $delta>0$ for every $epsilon>0$ such that
$$|f(g(x))-L|<epsilonhspace{0.25in}text{whenever}hspace{0.25in}0<|x-b|<delta$$
Since $displaystylelim_{xto a}f(x)=L$, there exists a $delta_1>0$ such that
$$|f(x)-L|<epsilonhspace{0.25in}text{whenever}hspace{0.25in}0<|x-a|<delta_1$$
And, since $displaystylelim_{xto b}g(x)=a$, then we can choose $delta$ in the following manner:
$$|g(x)-a|<delta_1hspace{0.25in}text{whenever}hspace{0.25in}0<|x-b|<delta$$
The proof goes like this: $0<|x-a|<delta_1Rightarrow|f(x)-L|<epsilon$ only means that $x$ is a number that's within a distance of $delta_1$ from $x=a$ such that plugging it into $f(x)$ gives you a number that's within a distance of $epsilon$ from $L$. So assuming that $0<|x-b|<delta$, it's true that $|g(x)-b|<delta_1$, which means that you can plug $g(x)$ in $f(x)$ and it'll return a number that's within a distance of $epsilon$ from $L$. Therefore, $|f(g(x))-L|<epsilon$.

However, this proof forgets that $g(x)$ can equal $a$, even when $xne b$. When that happens, plugging that to $f(x)$, which we have not specified to be continuous at $x=a$, would return an undefined result if otherwise. Therefore, following this proof, assuming $0<|x-b|<delta$ doesn't always assume $|f(g(x))-L|<epsilon$, as long as there exists an $xne b$ within $delta$ of $b$ such that $g(x)=a$. How can I make it work for all $x$ within $delta$ of $b$?

You can't make it work. The statement is false.

Take, for instance $f,gcolonmathbb{R}longrightarrowmathbb R$ defined by$$f(x)=begin{cases}1&text{ if }x=0\0&text{ otherwise}end{cases}text{ and }g(x)=begin{cases}0&text{ if }x=frac1ntext{ for some }ninmathbb N\x&text{ otherwise.}end{cases}$$Then:

• 
  $displaystyle lim_{xto0}f(x)=0$;


• 
  $displaystyle lim_{xto0}g(x)=0$;


• since $displaystyle fbigl(g(x)bigr)=begin{cases}1&text{ if }x=frac1ntext{ for some }ninmathbb N\0&text{ otherwise,}end{cases}$ the limit $displaystylelim_{xto0}fbigl(g(x)bigr)$ doesn't exist.

This would not happen if the definition of $displaystylelim_{xto a}f(x)=l$ was$$(forallvarepsilon>0)(existsdelta>0):lvert x-arvert<deltaimpliesbigllvert f(x)-lbigrrvert<varepsilon.$$