Prove: if $f(x)$ is continuous on $[a,b]$ where $x_o \in\left ]a,b\right[$ then,$\lim \limits_{x\to x_o} f(g(x)) = f( \lim \limits_{x \to x_o}g(x))$

Question

If $f(x)$ is continuous function on $]a,b[$ where $x_o \in \left ]a,b\right[$

prove that: $\lim \limits_{x \to x_o} f(g(x)) = f \left( \lim \limits_{x \to x_o}g(x) \right)$

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This was used in calculating a limit, in that limit, $f(x)=\ln(x)$. I think this is true for any continuous $f(x)$, so I look for any source provide a proof for this if it's true always.

Answer 1

The limit is the unique constant c such that $\forall \epsilon>0 \exists \delta \ s.t.f(g(x_0+\delta))-c<\epsilon$. since $f,g $ are continuous the limt exists and is the same from both sides, and $\forall \epsilon>0 \exists \delta \ s.t.f(g(x_0)+\delta)-f(g(x_0))<\epsilon$, and $\forall \delta \exists \delta' \ s.t.f(g(x_0+\delta'))-f(g(x_0))<|\delta|$. Hence $c=f(g(x_0))$ and since $\displaystyle\lim_{x\to x_0} g(x)=g(x_0), c=f(\lim_{x\to x_0}g(x))$

Answer 2

Let $k=\lim_{x\to x_0}g(x)$ and assume that $f$ is continuous at $k$. Then we want to show that: $$ \lim_{x\to x_0}f(g(x))=f(k) $$ which is equivalent to showing that:

$$ \forall \epsilon > 0,~\exists \delta > 0 \quad\text{ such that }\quad 0<|x-x_0| < \delta \implies |f(g(x))-f(k)|<\epsilon $$

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Note that since $f$ is continuous at $k$, we have $\lim_{y\to k}f(y)=f(k)$, so we know that: $$ \forall \epsilon_1 > 0,~\exists \delta_1 > 0 \quad\text{ such that }\quad |y-k| < \delta_1 \implies |f(y)-f(k)|<\epsilon_1 \tag{1} $$ Now since $\lim_{x\to x_0}g(x)=k$, we know that: $$ \forall \epsilon_2 > 0,~\exists \delta_2 > 0 \quad\text{ such that }\quad 0<|x-x_0| < \delta_2 \implies |g(x)-k|<\epsilon_2 \tag{2} $$ With that in mind, choose any $\epsilon>0$. Now let $\epsilon_1=\epsilon$. Using $(1)$, choose $\epsilon_2$ to be the $\delta_1$ corresponding to $\epsilon_1=\epsilon$. Using $(2)$, choose $\delta$ to be the $\delta_2$ corresponding to $\epsilon_2=\delta_1$.

Observe that if $0<|x-x_0|<\delta=\delta_2$, then we know from $(2)$ that $|g(x)-k|<\epsilon_2=\delta_1$. But then from $(1)$, we know by letting $y=g(x)$ that $|f(g(x))-f(k)|<\epsilon_1=\epsilon$, as desired.