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Prove that $\begin{equation*} \lim_{x \rightarrow 1} |x + 2|=3 \end{equation*}$ using limit definition>

I have a problem when I substitute for $|x + 2|$ by $-x-2$, could anyone help me in this case please? I could not reach $|x-1|$ which is less than $\delta$ in this case.

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The crucial trick is that: $\bigg||x+2|-3\bigg|\leq|(x+2)-3|=|x-1|$ by triangle inequality.

@Michael Hardy has demonstrated that triangle inequality is not needed by looking locally at $x=1$, personally I think one should choose $\delta=\min\{1,\epsilon\}$ to assure that $x$ does not vary too far away to get the positiveness of $x+2$.

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  • $\begingroup$ How this is triangle inequality? could you clarify please? $\endgroup$ – user426277 Nov 27 '17 at 1:15
  • $\begingroup$ $|a|=|(a-b)+b|\leq|a-b|+|b|$, so $|a|-|b|\leq|a-b|$. By symmetry, $|b|-|a|\leq|b-a|$, but $|b-a|=|a-b|$, so $\bigg||a|-|b|\bigg|\leq|a-b|$. $\endgroup$ – user284331 Nov 27 '17 at 1:17
  • $\begingroup$ You don't need the triangle inequality. When $x$ is near $1,$ then $x+2$ is positive, so $|x+2|$ reduces to $x+2. \qquad$ $\endgroup$ – Michael Hardy Nov 27 '17 at 1:17
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    $\begingroup$ Yes, true, luckily my solution still works. $\endgroup$ – user284331 Nov 27 '17 at 1:18
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When $x$ is near $1,$ then $x+2$ is positive, so $|x+2|$ would simplify to just $x+2,$ not $-x-2.$

So if $|x-1|<\delta$ then $|f(x) - 3| = |(x+2) - 3| = |x-1|<\delta.$ Thus setting $\delta=\varepsilon$ suffices.

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