I've been playing around with the limits of compositions…e.g. $\displaystyle \lim_{x \to a}(f \circ g) (x)$…and wanted to confirm my intuition that just because $\displaystyle \lim_{x \to a}(f \circ g) (x)=L$, it is not necessarily the case that $\displaystyle \lim_{x \to g(a)}f(x)=L$.
Said differently, if $\displaystyle \lim_{x \to a}(f \circ g) (x)=L$, then we cannot generally conclude that $\displaystyle \lim_{x \to g(a)}f(x)=L$.
The pictures (hand drawn, sorry) I had in mind are the following:
where the function $g$ maintains its value for all $x \geq a$.
The definition of $\displaystyle \lim_{x \to a}(f \circ g) (x)=L$, which is $\forall \epsilon \gt 0 \exists \delta \gt 0 \forall x \in \mathbb R \big [ 0 \lt \lvert x -a \rvert \lt \delta \rightarrow \lvert (f \circ g)(x) – L \rvert \lt \epsilon \big ] $ for $L = f(g(a))$, seems to be satisfied. From the graph of $f$, clearly $\displaystyle \lim_{x \to g(a)}f(x)\neq f(g(a))$.
Is this the correct intuition?
Best Answer
To grasp the concept, just take $g(x)=G$ a constant function.
When determining the limit $\lim\limits_{x\to a}f(g(x))=\lim\limits_{x\to a}f(G)=f(G)$ you can see clearly that the neighbourhood of $a$ is not even involved since $g$ is not varying.
It is therefore impossible to deduct the local behaviour of $f$ from a single value.
It suffice then to take $f$ discontinuous at $G$ so that $\lim\limits_{x\to G}f(x)$ is not defined.
This is an extreme example, but it illustrates where the problem originates.
Similarly imagine you have $g$ positive and want to study the limit of $f$ at $0$. The problem is despite $g$ being well behaved (continuous and all) since it takes only positive values, at most you can only get information for $f$ in $0^+$ and there is no way to know what happen on the left side in $0^-$ and furthermore determine the status of $\lim_{x\to 0}f(x)$.