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zenterix

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- Homework Statement
- I'm doing all the problems in Calculus by Spivak. I am on problem 3 of chapter 5 on Limits, and some of the items don't have solutions. I am interested in two particular items of this problem, ##vii## and ##viii##

- Relevant Equations
- We are asked to determine the limit ##l## for the given function and value of ##a##, and to prove that it is the limit by showing how to find a ##\delta## such that ##|f(x)-l|<\epsilon## for all ##x## satisfying ##0<|x-a|<\delta##.

The solution for this problem is not available in the solution manual.

Consider item ##vii##, which specifies the function ##f(x)=\sqrt{|x|}## with ##a=0##

Case 1: ##\forall \epsilon: 0<\epsilon<1##

$$\implies \epsilon^2<\epsilon<1$$

$$|x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Case 2: ##\forall \epsilon: 1\leq \epsilon < \infty##

$$\epsilon\leq\epsilon^2 \implies |x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Therefore

$$\forall \epsilon>0,|x-0|<\epsilon^2\implies|\sqrt{|x|}-0|<\epsilon$$

$$\implies \lim_\limits{x\to a} \sqrt{|x|}=0$$

$$|x-1|=|(\sqrt{x})^2-1^2|=|(\sqrt{x}+1)(\sqrt{x}-1)|\leq|\sqrt{x}+1||\sqrt{x}-1|$$

Assume ##|\sqrt{x}-1|<1##

$$\iff -1<\sqrt{x}-1<1\implies 1<\sqrt{x}+1<3$$

$$\iff |\sqrt{x}+1|<3$$

$$\iff |x-1|<3|\sqrt{x}-1|<3\epsilon$$

In words: Assume that ##f(x)## is less than 1 unit away from the value ##1##. From this follows that ##|x-1|<3|\sqrt{x}-1|<3\epsilon##. However, because all the steps involved an ##\iff##, we could actually start from assuming ##|x-1|<3|\sqrt{x}-1|<3\epsilon## and reach ##|\sqrt{x}-1|<1##.

In any case the conclusion is that

$$\forall \epsilon>0,|x-1|<3\epsilon\implies |\sqrt{x}-1|<min(1,\epsilon)<\epsilon$$

$$\implies \lim_\limits{x \to 1}\sqrt{x}=1$$

Case 1: ##\forall \epsilon: 0<\epsilon<1##

$$\implies \epsilon^2<\epsilon<1$$

$$|x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Case 2: ##\forall \epsilon: 1\leq \epsilon < \infty##

$$\epsilon\leq\epsilon^2 \implies |x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Therefore

$$\forall \epsilon>0,|x-0|<\epsilon^2\implies|\sqrt{|x|}-0|<\epsilon$$

$$\implies \lim_\limits{x\to a} \sqrt{|x|}=0$$

**Is this correct?**Now consider item ##viii##, which specifies the function ##f(x)=\sqrt{x}## near ##x=1##$$|x-1|=|(\sqrt{x})^2-1^2|=|(\sqrt{x}+1)(\sqrt{x}-1)|\leq|\sqrt{x}+1||\sqrt{x}-1|$$

Assume ##|\sqrt{x}-1|<1##

$$\iff -1<\sqrt{x}-1<1\implies 1<\sqrt{x}+1<3$$

$$\iff |\sqrt{x}+1|<3$$

$$\iff |x-1|<3|\sqrt{x}-1|<3\epsilon$$

In words: Assume that ##f(x)## is less than 1 unit away from the value ##1##. From this follows that ##|x-1|<3|\sqrt{x}-1|<3\epsilon##. However, because all the steps involved an ##\iff##, we could actually start from assuming ##|x-1|<3|\sqrt{x}-1|<3\epsilon## and reach ##|\sqrt{x}-1|<1##.

In any case the conclusion is that

$$\forall \epsilon>0,|x-1|<3\epsilon\implies |\sqrt{x}-1|<min(1,\epsilon)<\epsilon$$

$$\implies \lim_\limits{x \to 1}\sqrt{x}=1$$

**Is this correct?**