Question

Prove that limx->2 x^2=4

Answer 1

ok i know how to do this but

Answer 2

=2^2=4

Answer 3

we can use it as a approximation

Answer 4

no epsilon delta proof

Answer 5

\[(x-2)(x+2)<\epsilon\]

Answer 6

\[|(x-2)|(x+2)|<\epsilon\]\[|x-2|<\delta\]\[-\delta<|x-2|<\delta\]\[4-\delta<x+2<\delta+4\]
so we use the lesser value right?

Answer 7

to plug in to solve for delta?

Answer 8

\[\delta(4+\delta)<\epsilon\]

Answer 9

Isn't it easier to prove by showing that as x approaches 2, the value approaches 4?
So, you can use x = 1.8, 1.9, 1.99, 1.9999.
This should prove the limit as x approaches 2, x^2 becomes 4

Answer 10

Let \[\varepsilon > 0.\]We must find some\[\delta>0\]such that\[|x-2|<\delta\Rightarrow|x^2 - 4|<\varepsilon.\]
\[|x^2-4|=|x-2||x+2|<\delta|x+2|.\]Since \[-\delta<x-2<\delta \Rightarrow 4 - \delta<x+2<\delta+4.\]If \[\delta<2\]then \[x + 2 > 0\]so\[|x+2| = x+2<\delta + 4\]and\[|x^2 - 4| < \delta(\delta + 4).\]For \[0 < \delta < -2 + \sqrt{4+\varepsilon}\]we have\[\delta(\delta+4)<\varepsilon\]so we for any \[\delta\]that verifies\[0 < \delta < \min\{2, -2 + \sqrt{4 + \varepsilon}\}\]we have\[|x - 2| < \delta \Rightarrow |x^2 - 4| < \varepsilon\]and the proof is complete.

Answer 11

ok thank you i know how to do it i just wanted to know if we use the lesser value but nvm