Let us define two functions:-
$$f(x)$$
$$g(x)$$
And now let us try to find the limit :-
$$\lim_{x \to c} f(g(x))$$
Assuming that all the limits mentioned below exist we can say:-
$$ \lim_{x \to c}g(x)=L$$
$$\lim_{x \to c}f(g(x))=\lim_{g(x) \to L} f(\lim_{x \to c}g(x))=\lim_{x \to L}f(x)$$
That is is just taking the limit of $f$ when the input is limit of $g$.
But why is it so why should we use the limit of $g$ as the input for $f$ according to me it is not equivalent to:-
$$\lim_{x\to c}(f \circ g)x$$
So why do we use it?
Edit- After seeing the comments I think people do not understand what I am saying are not equivalent. I want to say according to me:-
$$\lim_{x\to c}(f \circ g)x ≠ \lim_{x \to L}f(x)$$
So why do we use it to find limits of composite functions?
Best Answer
It is not true that if $\lim_{x\to c} g(x) = L$ and $\lim_{x \to L} f(x) = u$ then $\lim_{x \to c} f(g(x)) = u$. Here is a counterexample. Let $$ f(x) = \begin{cases} 4, &\text{if } x \ne 7,\\ 3, &\text{if } x = 7, \end{cases} $$ and $g(x) = 7$. Then $\lim_{x \to 0} g(x) = 7$ and $\lim_{x \to 7} f(x) = 4$, but for all $x$, $f(g(x)) = f(7) = 3$, so $\lim_{x \to 0} f(g(x)) = 3$.
Intuitively, here is the problem. To simplify notation, let $y$ denote $g(x)$, so $f(g(x)) = f(y)$. And let us rewrite $\lim_{x \to L} f(x)$ as $\lim_{y \to L} f(y)$ (they are the same). Then $\lim_{x \to c} g(x) = L$ means that if $x$ is close to $c$ but not equal to $c$ then $y = g(x)$ is close to $L$. And $\lim_{y \to L} f(y) = u$ means that if $y$ is close to $L$ but not equal to $L$ then $f(y)$ is close to $u$. The problem when trying to combine these is that if $x$ is close to $c$ but not equal to $c$, we know that $y$ is close to $L$, but we don't know that $y$ is not equal to $L$, so we can't apply $\lim_{y \to L} f(y) = u$.
In my calculus book (Calculus: A Rigorous First Course), I introduced the following notation. If $y = g(x)$ and $\lim_{x \to c} g(x) = L$, many books would write "as $x \to c$, $y \to L$," but this notation leaves out the fact that $x$ can't be equal to $c$. So in my book, I write "as $x \to c^{\ne}$, $y \to L$." Using this notation, if $y = g(x)$, $\lim_{x \to c} g(x) = L$, and $\lim_{y \to L} f(y) = u$, then we have: as $x \to c^{\ne}$, $y \to L$ and as $y \to L^{\ne}$, $f(y) \to u$. But the mismatch between $y \to L^{\ne}$ and $y \to L$ means that these statements cannot be combined.
If you know that as $x \to c^{\ne}$, $y = g(x) \to L$ and as $y \to L$, $f(y) \to u$, then you can combine these statements and say that as $x \to c^{\ne}$, $f(g(x)) = f(y) \to u$. This is the justification for the theorem, found in most calculus books, that if $\lim_{x \to c} g(x) = L$, $f$ is continuous at $L$, and $f(L) = u$, then $\lim_{x \to c} f(g(x)) = u$.
If you know that as $x \to c^{\ne}$, $y = g(x) \to L^{\ne}$ and as $y \to L^{\ne}$, $f(y) \to u$, then you can also conclude that as $x \to c^{\ne}$, $f(g(x)) = f(y) \to u$, so $\lim_{x \to c} f(g(x)) = u$. This shows that if $\lim_{x \to c} g(x) = L$, $\lim_{y \to L} f(y) = u$, and for $x \ne c$, $g(x) \ne L$, then you can conclude that $\lim_{x \to c} f(g(x)) = u$.