Solution:
Given \( \epsilon >0 \), we want to find \( \delta \) , such that
\( 0<|x-4|< \delta \implies |\sqrt{x}-2|< \epsilon \).
So for \( 0<|x-4|< \delta \)
\( - \delta < x-4 < \delta , x \neq 4\)
\( 4-\delta < x < 4+ \delta \)
And for \( |\sqrt{x}-2| < \epsilon \)
\( - \epsilon < \sqrt{x}-2 < \epsilon \)
\( 2- \epsilon < \sqrt{x} < 2 + \epsilon \)
\( (2- \epsilon )^2 < x < (2+ \epsilon)^2 \)
So we need to find \( \delta \) such that
\( (2- \epsilon)^2< 4 - \delta < x < 4+ \delta < (2+ \epsilon)^2 \)
So we need \( (2- \epsilon)^2 < 4 -\delta \) and \( 4+ \delta < (2+ \epsilon)^2 \)
\( \delta < 4-(2- \epsilon)^2 \) and \( \delta< (2+ \epsilon)^2 -4 \)
So we need the smaller \( \delta \), so we take \( \delta < 4-(2- \epsilon )^2 \)
