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As per the definition of limits if $\lim_{x \to a} f(x)= L$, then $$\forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ s.t 0\lt\lvert x-a \rvert \lt \delta \ \implies \ 0\lt \lvert f(x)- L\rvert \lt \varepsilon $$

I want to prove that $\lim_{x \to a} x^2 = a^2$. As per the definition $$\lvert f(x)- L\rvert = \lvert x^2- a^2\rvert = \lvert (x-a)(x+a)\rvert =\lvert x-a\rvert \lvert x+a \rvert$$

As per definition $$\lvert x-a \rvert \lt \delta \implies -\delta \lt x-a \lt \delta \implies a-\delta \lt x <a+\delta \implies 2a-\delta \lt x+a <2a+\delta $$

I'm stuck beyond this. I cannot find a suitable $\varepsilon$ to satisfy my condition here.

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  • $\begingroup$ take $\delta=\min(|a|,\epsilon)$ you get $|x^2-a^2|<3|a|\epsilon$ $\endgroup$
    – zwim
    Commented Nov 10, 2021 at 19:55
  • $\begingroup$ I think the idea put forward by the OP is a good one. Why do the solutions seems to go off the deep end? $\endgroup$
    – user798113
    Commented Nov 10, 2021 at 20:23
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    $\begingroup$ Note that you have $2a-1<x+a<2a+1$ so we have that $|x+a|<2|a|+1$ (since $|x+a|=|(x-a)+2a|\leqslant |x-a|+2|a|$) and then $|x^{2}-a^{2}|=|x-a||x+a|=(2|a|+1)\delta$. Now you can choose $\delta(\varepsilon)=\min\left\{1,\frac{\varepsilon}{2|a|+1}\right\}$. $\endgroup$
    – user798113
    Commented Nov 10, 2021 at 20:28

4 Answers 4

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A different approach for the sake of curiosity.

Let $0 < |x - a| < \delta_{\varepsilon}$. Then we have that: \begin{align*} |f(x) - L| & = |x^{2} - a^{2}|\\\\ & = |(x-a)(x+a)|\\\\ & = |x - a||(x - a) + 2a|\\\\ & \leq |x - a|(|x - a| + 2|a|)\\\\ & < \delta_{\varepsilon}(\delta_{\varepsilon} + 2|a|) := \varepsilon \end{align*}

where you can choose the positive root of the corresponding equation on $\delta_{\varepsilon}$.

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    $\begingroup$ you are saying that I frame quadratic roots in terms of a and $\varepsilon$? $\endgroup$
    – Orpheus
    Commented Nov 10, 2021 at 20:16
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    $\begingroup$ That's the point. Given a certain $\varepsilon > 0$, you can choose $$\delta_{\varepsilon} = -|a| + \sqrt{a^{2} + \varepsilon}$$ so that the corresponding definition is satisfied. $\endgroup$ Commented Nov 10, 2021 at 21:50
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    $\begingroup$ Thanks a lot @Átila Correia $\endgroup$
    – Orpheus
    Commented Nov 11, 2021 at 3:36
  • $\begingroup$ This is definitely the best answer, as this approach is by far the simplest and most intuitive. $\endgroup$
    – Angel
    Commented Nov 11, 2021 at 4:05
  • $\begingroup$ @Orpheus You are welcome ! I am glad I could help. $\endgroup$ Commented Nov 11, 2021 at 16:03
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Alternative approach:

Without loss of generality, $a > 0$. That is, the approach for $a < 0$ is similar, while the approach if $a = 0$ is trivial.

If $\delta = \epsilon,$ then $|x-a| < \delta \implies |x-a| < \epsilon$.

Instead, take
$\displaystyle \delta = \min\left(\frac{a}{2},\frac{\epsilon}{3a}\right)$.

Then, $|x - a| < \delta \implies \frac{a}{2} < x < \frac{3a}{2}$.

This implies that $|x + a| < \frac{5a}{2} < 3a$.

So, you have that $|x - a| < \delta$ and
$|x + a| < 3a$.

Therefore

$|x^2 - a^2| = |x - a| \times |x + a| < \delta \times 3a \leq \epsilon$.

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  • $\begingroup$ I like your approach....but why did you take $\delta = \frac{a}{2}$ in the begining? $\endgroup$
    – Orpheus
    Commented Nov 11, 2021 at 2:48
  • $\begingroup$ @Orpheus, I took $\displaystyle\delta = \min\left(\frac{a}{2}, \frac{\epsilon}{3a}\right)$ so that I could infer that $\displaystyle x < \frac{3a}{2} \implies (x + a) < \frac{5a}{2} < 3a.$ $\endgroup$ Commented Nov 11, 2021 at 3:12
  • $\begingroup$ This is just a question....so I'm allowed to pick any $\delta$? $\endgroup$
    – Orpheus
    Commented Nov 11, 2021 at 3:38
  • $\begingroup$ $\delta$ has to (somehow) be based on $\epsilon$. However, additional constraints can be assigned to $\delta$, that are based on fixed values. For example, with $a$ assumed to be a fixed positive number, you can force $\delta \leq \frac{a}{2}$. Similarly, in other problems, if you want to guarantee that $\delta^2 < \delta$, you can force $\delta \leq 1.$ $\endgroup$ Commented Nov 11, 2021 at 3:42
  • $\begingroup$ So for this problem, I could try a different $\delta$? $\endgroup$
    – Orpheus
    Commented Nov 11, 2021 at 3:50
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For all real numbers $x$ and $a$ we have \begin{eqnarray} |x^2-a^2|&=&|(x-a)(x+a)|\cr &=&|x-a||x+a|\cr &=&|x-a||(x-a)+a|\cr &\le&|x-a|(|x-a|+|a|)\cr &=& |x-a|^2+|a||x-a| \end{eqnarray} If $\varepsilon>0$ is such that

$$ |x-a|^2+|a||x-a|<\varepsilon, $$

then $$ \frac{-|a|-\sqrt{a^2+4\varepsilon}}{2}<|x-a|<\frac{-|a|+\sqrt{a^2+4\varepsilon}}{2} $$

Hence, if we choose $$ \delta=\frac{-|a|+\sqrt{a^2+4\varepsilon}}{2}=\frac{2\varepsilon}{|a|+\sqrt{a^2+4\varepsilon}}>0 $$ for all $x$ satisfying $|x-a|<\delta$, we have $|x^2-a^2|<\varepsilon$

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  • $\begingroup$ Thanks a lot, @Mercy King $\endgroup$
    – Orpheus
    Commented Nov 11, 2021 at 3:39
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You must find a solution to the equation in $\delta$

$$|x-a|<\delta\implies|x^2-a^2|<\epsilon.$$

The RHS is

$$\sqrt{a^2-\epsilon}<x<\sqrt{a^2+\epsilon}$$ or

$$\sqrt{a^2-\epsilon}-a<x-a<\sqrt{a^2+\epsilon}-a.$$

So if we take

$$0<\delta<\min(a-\sqrt{a^2-\epsilon},\sqrt{a^2+\epsilon}-a)$$ the conditions are fulfilled.

Note that if $\epsilon\ge a^2$, the condition $a^2-x^2<\epsilon$ is automatically fulfilled by $0\le x\le a$.

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  • $\begingroup$ thanks a lot, @Pat Bol $\endgroup$
    – Orpheus
    Commented Nov 11, 2021 at 3:40

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