Question

Use the epsilon-delta definition to prove that the limit is L

Answer 1

\[\lim_{x \rightarrow 4} \sqrt{x}=2\]

I've already done this:
\[|x-4|< \delta\]
\[|\sqrt{x}-2|<\epsilon\] \[|(\sqrt{x}-2)(\sqrt{x}+2)|<\epsilon|\sqrt{x}+2|\]
\[|x-4|<\epsilon|\sqrt{x}+2|\]

How do I continue?

Answer 2

\[|x|<4+\delta \]

Answer 3

\[|\sqrt{x}-2|<\epsilon \implies -\epsilon < \sqrt{x}-2<\epsilon \]

Answer 4

\[\implies -\epsilon +2 < \sqrt{x} < \epsilon +2 \implies (2-\epsilon) ^2<x < (2+\epsilon)^2 \]

Answer 5

\[\implies (2-\epsilon)^2-4 <x-4 < (2+\epsilon )^2-4\]

Answer 6

\[\implies \epsilon^2-2\epsilon<x-4<\epsilon^2+2\epsilon \]

Answer 7

\[\implies |x-4|<\delta \]

Answer 8

\[where \delta=\min{( \epsilon^2-2\epsilon, \epsilon^2+2\epsilon)}\]

Answer 9

@prepfortears

Answer 10

done