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Show the following statement is correct using the \(\epsilon\) - \(\delta\) definition of limit. \[\lim_{x \rightarrow2}(x^2 + x - 1) = 5\]
So, as a reminder, the epsilon delta defintion of a limit \[ \lim_{x \rightarrow a} f(x) = L\] \[\forall~\epsilon>0 ~~\exists~\delta>0 \] Then\[0<|x-a|<\delta \implies |f(x)-L|<\epsilon \]
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