Question

proove that limit x approaches 4 ,square root of x is equal to 2, using epilson delta formula.

Answer 1

\[
\lim_{x \rightarrow 4}\sqrt{x} = 2 \\
f(x) = \sqrt{x} \\
\text{For }|x-4| < \delta, \text{ find an } \epsilon \text{ such that } |\sqrt{x} - 2| \lt \epsilon \\
|x-4| < \delta \\
|\sqrt{x}+2||\sqrt{x}-2| \lt \delta ~~~\text{ ---- (1) }\\
\text{Lowest value x can have is 0 because f(x) is valid only for } x \ge 0 \\
\text{Lowest value } |\sqrt{x}+2| \text{ can have is 2 because } x \ge 0 \\
\text{From (1): } ~~ |\sqrt{x}-2| \lt \delta \\
\text{Let } \epsilon = \delta \\
|\sqrt{x}-2| \lt \epsilon \quad \Rightarrow \quad |f(x) - 2| \lt \epsilon\\
\text{Therefore, the limit is 2.}
\]

Answer 2

nice proof

Answer 3

@aum
Since \(|\sqrt x+2|\ge2\), we know \(\dfrac{1}{|\sqrt x+2|}\le\dfrac{1}{2}\), so
\[|\sqrt x-2|=\frac{|x-4|}{|\sqrt x+2|}\le\frac{1}{2}|x-4|<\epsilon~~\implies~~|x-4|<2\epsilon=\delta\]

Answer 4

Yes. I think either proof should be acceptable whether \(\epsilon = \delta \) or \(\epsilon = \frac 12\delta\).
\(|\sqrt{x}+2||\sqrt{x}-2| \lt \delta ~~~\text{ ---- (1) }\\\)
Since \(|\sqrt{x}+2| \ge 2\), I could have said from (1): \(|\sqrt{x}-2| \lt \frac 12\delta\) and so \(\epsilon = \frac 12\delta\).
But for simplicity I said: If (a number greater than 2) times A < B then certainly A < B without having to refine B dwon to B/2.

Answer 5

thank you all for your help.

Answer 6

i have have understood this proove. But in my course book they choose\[\delta =\min {4,\epsilon} \]. why?

Answer 7

They are just slightly different approaches to proving the limit. The important thing is in all cases, as \(\delta \rightarrow 0\), \(\epsilon \rightarrow 0\) thus proving the limit.

Answer 8

thanks @ aum