Question

use the epsilon - little delta definition to prove that the lim as x approaches 2 (6x^2 - 3) =21

Answer 1

Prove that
\[ \large \lim_{x\to2}(6x^2-3)=21. \]

Let \(\varepsilon>0\) then
\[ \large |6x^2-3-21|=|6x^2-24|=6|x^2-4|=6|x-2|\cdot|x+2|. \]
Let \(\delta=1\) so
\[ \large |x-2|<\delta=1 \]
\[ \large -1<x-2<1 \]
\[ \large 1<x<3 \]
\[ \large 3<x+2<5\qquad\text{so}\qquad |x+2|<5 \]
Then
\[ \large |6x^2-24|=6|x-2|\cdot|x+2|<6\cdot5|x-2|<\varepsilon \]
\[ \large \Rightarrow |x-2|<\varepsilon/30. \]
Therefore \(\delta=\min\{1,\varepsilon/30\}\).

Answer 2

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Answer 3

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