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Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to 0} | x | = 0 $

Given $\epsilon>0,$ we need $\delta>0$ such that if $0<|x-0|<\delta$, then ||$x|-0|<\epsilon .$ But ||$x||=|x| .$ So this is true if we pick $\delta=c$

Thus, $\lim _{x \rightarrow 0}|x|=0$ by the definition of a limit.

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there's a problem. Number twenty seven of the Stuart Calculus eighth edition, section two point four. Proof this demon using the Amazon Delta definition of a limit. The limit is expert zero of the function Absolutely. Vex is equal to zero. The absolute dealt the absolute delta definition of a limit states that this is, ah, true statement. This limit is equal to zero as long as we can find it. Have someone greater than zero as long as we can find a delta green zero for every Absalon greater than zero such that if the absolute value of the difference between accent A is lesson doctor, then the absolutely the difference between a function and l will be less turn. Absalon, let us work with the second inequality and play in our own function. Absolutely backs and subtract the limit which is equal to zero. And this is less than absolute. Here we're left with the apse of value of the absolute value of X. Ah, only one of these is necessary and we concluding this statement absolutely of exit lesson Absalon. If we begin with the other inequality, we write it as absolute value of X minus aids AIDS equals zero. Listen, Delta and we simplified Teo just after five axis less than Delta. And then we have to roll nearly identical statements. They're only identical if Delta is equal to Absalon. And this is an example for a choice of Delta that proves that you can find any Delta given any absolute any Absalon greater than zero. Thus, by the definition of the limit Absalon, Delta Definitional this limit. We have proven that this limit exists and is equal to zero.