Epsilon delta proof of nested sine function

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 $$|f(\sin(x))-\pi|<\epsilon$$?

• In your first sentence, did you mean, I know that $\lim_{x->0}f(x)=\pi$ and $\lim_{x->0}\sin(x)=0\$ ? Mar 7, 2022 at 14:42
• I think what I wrote is right Mar 7, 2022 at 16:05
• $\lim_{x->\pi}\sin(x)=0$ is true but irrelevant to the question. $\lim_{x->0}\sin(x)=0$ is relevant to the question. Mar 7, 2022 at 16:07

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.

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.

• I think the part that confused me the most was "Since sin is a function that outputs real values, this implies that: $\sin x\in (-\delta, \delta)\implies f(\sin x)-\pi\in (-\varepsilon, \varepsilon).$ Is that always true? Mar 7, 2022 at 16:21
• What makes you think it isn't true? Mar 7, 2022 at 17:12
• Updated my answer. Does it make any more sense now? Mar 8, 2022 at 13:33