Is the format of this proof for this limit correct?
I wasn't entirely sure what to make delta so I just made it a value that would result in parts of the final statement to cancel out.
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$\begingroup$This line is bothersome $|x-1|<1\implies|x+1|<3\implies|\frac 1{x^2+2}|<\frac 12$
It should be split into two statements both having $|x-1|<1$ as a premise.
In fact $|x-1|<1\implies x>0\implies |\frac 1{x^2+2}|<\frac 12$
But $|x+1|<3$ means $x\in]-4,2[$, in particular $x=0$ is possible, and in that case the inequality become loose $|\frac 1{x^2+2}|\le\frac 12$.
So you wrote a wrong implication, by chaining statements like this.
Also there is no reason to exclude $0$ in your statement $0<|x-1|<\delta$, it is just $|x-1|<\delta$.
The rest is OK.
The choice of $\delta=M$ is in general judicious for the rough bounding from above, and the choice $\delta=k\varepsilon$ is necessary for the continuity. Thus it is indeed pretty standard to end up with $\delta=\min(M,k\varepsilon)$ for $\varepsilon,\delta$ proofs.
Note: it is not mandatory to get precisely $1\times\varepsilon$ at the end, till you get $|f(x)-f(x_0)|<C\varepsilon$ with a constant $C$, then it is fine. So you could have taken $\delta=\min(1,\varepsilon)$ as well.
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