Epsilon delta proof of nested sine function

Question

I know that $\lim_{x->0}f(x)=\pi$ and $\lim_{x->\pi}\sin(x)=0$.
I have to prove that $\lim_{x->\pi}f(\sin(x))=\pi$ using the epsilon delta proof.
My idea was to use the value of $\delta=MAX(\delta_1,\delta_2)$ where $0<|x|<\delta_1$ and $0<|x-\pi|<\delta_2$. Then $|f(x)-\pi|<\epsilon$ and $|\sin(⁡x)|<\epsilon$.
Let $\epsilon>0$ and take $\delta=MAX(\delta_1,\delta_2)$. Assume $0<|\sin(x)-\pi|<\delta$. Then $|f(x)-\pi|<\epsilon$ and $|\sin(⁡x)|<\epsilon$. Can I from that somehow conclude that $|f(\sin(x))-\pi|<\epsilon$?

Answer 1

Let $\varepsilon>0$ be fixed. Since $f(x)\to\pi$ as $x\to 0$, there is some $\delta'>0$ such that

$$\left|f(x)-\pi\right|<\varepsilon$$

whenever $\left|x\right|<\delta'$. Similarly, since $\sin(x)\to0$ as $x\to\pi$, there is some $\delta>0$ such that

$$\left|\sin(x)\right|<\delta'$$

whenever $\left|x-\pi\right|<\delta$. Combining these we notice that we have the implications

$$\left|x-\pi\right|<\delta \implies \left|\sin(x)\right|<\delta' \implies \left|f(\sin(x))-\pi\right|<\varepsilon.$$

Thus

$$\left|f(\sin(x))-\pi\right|<\varepsilon$$

whenever $\left|x-\pi\right|<\delta$, which proves the assertion.

Answer 2

Assuming what was meant was: We are given that $\lim_{x\to0}f(x)=\pi$ and $\lim_{x\to0}\sin(x)=0.$

$\lim_{x\to0}f(x)=\pi\ $ means that: given $\varepsilon>0,\ \exists\ \delta>0$ such that $x\in (-\delta, \delta)\implies f(x)-\pi\in (-\varepsilon, \varepsilon).\ (1)$

$\lim_{x\to0}\sin(x)=0\ $ means that: given $\varepsilon'>0,\ \exists\ \delta'>0$ such that $x\in (-\delta', \delta')\implies \sin x\in (-\varepsilon', \varepsilon').\ (2)$

$$$$

Propoition: $\lim_{x\to0}f(\sin(x))=\pi.\ $ Proof: Let $\varepsilon>0.$ Then from $(1),$ there exists $\delta>0$ such that $u\in (-\delta, \delta)\implies f(u)-\pi\in (-\varepsilon, \varepsilon).$ Since $\sin$ is a function that inputs and outputs real values only, this implies that: $u=\sin x\in (-\delta, \delta)\implies f(\sin x)-\pi\in (-\varepsilon, \varepsilon).$ From $(2),$ there exists $\delta'>0$ such that $x\in (-\delta', \delta')\implies \sin x\in (-\delta, \delta),\ $ but from the previous sentence, this implies that $f(\sin x)-\pi\in (-\varepsilon, \varepsilon).$ This completes the proof.