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Please help me check the mistake in the following proof, thanks!

Prove: if $f$ is continuous at $x=1$ and $f(1)=7$, then $lim_{x→4}f(x_{2}−x−11)=7$ Proof. Fix $ϵ>0$. By continuity of $f$ at 1 , there exists $δ_{1}>0$, such that $∣y−1∣<δ_{1}$, which implies $∣f(y)−f(1)∣<ϵ$. So choose $δ=min{1,8δ_{1} }$, then $∣x+3∣∣x−4∣<δ$ implies $∣x−4∣<8δ_{1} =δ$. So $∣∣ (x_{2}−x−11)−1∣∣ <δ_{1}$, which implies $∣∣ f(x_{2}−x−11)−7∣∣ <ϵ$ if $∣x−4∣<8δ_{1} $. Therefore, if $f$ is continuous at $x=1$ and $f(1)=7$, then $lim_{x→4}f(x_{2}−x−11)=7$.

Analyzing the proof. Let's go step by step:

1. The statement you want to prove is: If f is continuous...