Given conditions: lim ⁡ ? → ? ? ( ? ) = ? lim t→a ​ g(t)=b lim ⁡ ? → ? ? ( ? ) = ? lim x→b ​ f(x)=c To prove: lim ⁡ ? → ? ? ( ? ( ? ) ) = ? lim t→a ​ f(g(t))=c

1. Given conditions:
   • lim⁡t→ag(t)=b\lim_{t \to a} g(t) = blimt→a​g(t)=b
   • lim⁡x→bf(x)=c\lim_{x \to b} f(x) = climx→b​f(x)=c
2. To prove:
   • lim⁡t→af(g(t))=c\lim_{t \to a} f(g(t)) = climt→a​f(g(t))=c
3. Using the definition of the limit:
   • lim⁡t→ag(t)=b\lim_{t \to a} g(t) = blimt→a​g(t)=b means that for every ϵ′>0\epsilon’ > 0ϵ′>0, there exists a δ′>0\delta’ > 0δ′>0 such that for all $t$ satisfying 0<∣t−a∣<δ′0 < |t – a| < \delta’0<∣t−a∣<δ′, we have ∣g(t)−b∣<ϵ′|g(t) – b| < \epsilon’∣g(t)−b∣<ϵ′.
   • lim⁡x→bf(x)=c\lim_{x \to b} f(x) = climx→b​f(x)=c means that for every $ϵ>0$, there exists a $\delta >0$ such that for all $x$ satisfying $0<|x-b|<\delta$, we have $|f(x)-c|<ϵ$.
4. Composition of limits:
   • To prove lim⁡t→af(g(t))=c\lim_{t \to a} f(g(t)) = climt→a​f(g(t))=c, we need to show that for every $ϵ>0$, there exists a $\delta >0$ such that for all $t$ satisfying $0<|t-a|<\delta$, we have $|f(g(t))-c|<ϵ$.
5. Combining the given limits:
   • Let $ϵ>0$ be given.
   • Since lim⁡x→bf(x)=c\lim_{x \to b} f(x) = climx→b​f(x)=c, there exists a $\delta >0$ such that for all $x$ satisfying $0<|x-b|<\delta$, we have $|f(x)-c|<ϵ$.
   • Now, since lim⁡t→ag(t)=b\lim_{t \to a} g(t) = blimt→a​g(t)=b, there exists a δ′>0\delta’ > 0δ′>0 such that for all $t$ satisfying 0<∣t−a∣<δ′0 < |t – a| < \delta’0<∣t−a∣<δ′, we have $|g(t)-b|<\delta$.
6. Putting it all together:
   • Choose δ′>0\delta’ > 0δ′>0 such that $|g(t)-b|<\delta$ whenever 0<∣t−a∣<δ′0 < |t – a| < \delta’0<∣t−a∣<δ′.
   • Now, if 0<∣t−a∣<δ′0 < |t – a| < \delta’0<∣t−a∣<δ′, then $|g(t)-b|<\delta$.
   • Since $|g(t)-b|<\delta$, by the limit of $f$, we have $|f(g(t))-c|<ϵ$.
7. Conclusion:
   • Thus, for every $ϵ>0$, we can find δ′>0\delta’ > 0δ′>0 such that if 0<∣t−a∣<δ′0 < |t – a| < \delta’0<∣t−a∣<δ′, then $|f(g(t))-c|<ϵ$.
   • This proves that lim⁡t→af(g(t))=c\lim_{t \to a} f(g(t)) = climt→a​f(g(t))=c.

Hence, we have shown that if lim⁡t→ag(t)=b\lim_{t \to a} g(t) = blimt→a​g(t)=b and lim⁡x→bf(x)=c\lim_{x \to b} f(x) = climx→b​f(x)=c, then lim⁡t→af(g(t))=c\lim_{t \to a} f(g(t)) = climt→a​f(g(t))=c