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What is the common proof for $\lim_{x\rightarrow c}x^2=c^2$. I tried like this

Let $\varepsilon>0$, want to find a $\delta>0$ such that $\forall x\in\mathbb{R},0<|x-c|<\delta\Rightarrow |x^2-c^2|<\varepsilon$. Therefore $$-\varepsilon<x^2-c^2<\varepsilon$$ $$c^2-\varepsilon<x^2<\varepsilon+c^2$$ $$-c-\sqrt{c^2+\varepsilon}<x-c<-c+\sqrt{c^2+\varepsilon}$$ Choose $\delta=-c+\sqrt{c^2+\varepsilon}$which proves the problem

Is this proof is correct ? If there is any mistake in it give a correct proof please .thanks

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    $\begingroup$ The usual proof would use the continuity of $x \mapsto x^2$. $\endgroup$ – Patrick Stevens Jul 24 '16 at 11:18
  • $\begingroup$ I won't understand can u explain it please $\endgroup$ – Sathasivam K Jul 24 '16 at 11:20
  • $\begingroup$ I think you have some mistake in getting from the second inequality to the third one! If $x^2<a$ then $|x|<\sqrt{a}$ or equivalently $-\sqrt{a}<x<\sqrt{a}$ $\endgroup$ – H. R. Jul 24 '16 at 11:47
  • $\begingroup$ Ya you are right ,then is the third inequality is like this?$$-c-\sqrt{c^2+\varepsilon}<x-c<-c+\sqrt{c^2+\varepsilon}$$ is this right? $\endgroup$ – Sathasivam K Jul 24 '16 at 11:52
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    $\begingroup$ @PatrickStevens Of course. My bad. $\endgroup$ – Aweygan Jul 24 '16 at 15:29
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You have made one common arithmetic mistake and two common conceptual mistakes.

The arithmetic and first conceptual mistake are mistakes of algebra. The second conceptual mistake is the first mistake that's actually related to calculus.


The arithmetic mistake is the usual one regarding square roots and forgetting that negative numbers exist — the detail here is a bit complicated; the easiest method would be to go back to your precalculus text and review solving inequalities.

Henceforth, I will assume all the variables are such that your work is correct.


The first conceptual mistake is the usual mistake of proving the converse of what you are trying to solve. You have proven

If $\lvert x^2 - c^2\rvert < \epsilon$, then $\lvert x-c\rvert < \delta$

but what you are trying to prove is

If $\lvert x - c\rvert < \delta$, then $\lvert x^2 - c^2\rvert < \epsilon$

This is a common mistake, because you're taught to solve problems backwards; working backwards is an effective means of simplifying problems, but once you have what you believe to be a solution, you have to turn around and prove that it is so. Again, you can see examples of this in your precalculus class, where you solve equations like $\sqrt{x+2} = x$ but get "spurious" solutions; in this case, $x=-1$.


The other conceptual mistake is that you are trying hard to solve the problem exactly. One of the great things that makes calculus "simple" is that we don't have to solve intervals like this exactly — we can usually make all sorts of simplifications to make our life easier.

The usual trick here is to observe that

$$ \lvert x^2 - c^2\rvert = \lvert x-c\rvert \lvert x+c\rvert $$

so if $\lvert x-c\rvert < \delta$, then

$$ \lvert x^2 - c^2 \rvert = \lvert x-c\rvert \lvert x+c\rvert < \delta \lvert x+c\rvert $$

To deal with the $\lvert x+c\rvert$ term, we simply observe that $x$ is near $c$, so $\lvert x+c\rvert$ shouldn't be much bigger than $\lvert 2c\rvert$.

So, we add a second goal: in addition to wanting to prove

$$ \lvert x - c\rvert < \delta \implies \lvert x^2 - c^2\rvert < \epsilon $$

we also want to have

$$ \lvert x - c\rvert < \delta \implies \lvert x + c\rvert < \lvert 2c\rvert + 1 $$

The latter isn't hard to get: it's enough to pick $\delta \leq 1$.

Thus, if we have both $\lvert x - c\rvert < \delta$ and $\delta < 1$, then we have

$$ \lvert x^2 - c^2\rvert = \lvert x-c\rvert \lvert x+c\rvert < \delta (\lvert 2c\rvert + 1) $$

So if we can additionally choose $\delta$ so that $\delta (\lvert 2c\rvert + 1) \leq \epsilon$, then we have what we need!

Choosing $\delta = \min\left( \frac{\epsilon}{\lvert 2c\rvert + 1}, 1 \right)$ works to satisfy both of the conditions we needed for $\delta$.

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    $\begingroup$ A great answer ....really it help me to rectify my mistakes $\endgroup$ – Sathasivam K Jul 24 '16 at 12:11
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First of all, your proof is not correct because if $c < 0$ then the inequality $\sqrt{c^2-\epsilon} < x$ is wrong since $x < 0$ as well. Here is my analysis:

There are $2$ cases:

  1. $c = 0$. Can you do this easy case?
  2. $c \neq 0$. Then you can force $\delta \le 1$, thus $|x-c| < 1 \implies |x+c| \leq |x-c| + |2c| < 1 + 2|c|$. Thus if you choose $\delta = \text{min}\left(1,\dfrac{\epsilon}{1+2|c|}\right)$, then you would have the good proof. Can you put it together?
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  • $\begingroup$ Can u explain it briefly sir? I am just a begginer and I join collage just a week ago $\endgroup$ – Sathasivam K Jul 24 '16 at 11:27
  • $\begingroup$ Sorry for distrubing you again sir ..but..you told..."if $c < 0$ then the inequality $\sqrt{c^2-\epsilon} < x$ is wrong since $x < 0$ as well. "...why x is need to be less than zero sir.since $\epsilon $ is very small $\sqrt{c^2-\epsilon} $ need not to be <0 then how the x became less thanzero? Moreover even if c<0,$c^2>0$. $\endgroup$ – Sathasivam K Jul 24 '16 at 11:44
  • $\begingroup$ Thanks for your answer sir@deep sea $\endgroup$ – Sathasivam K Jul 24 '16 at 11:54
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$$lim_{x \to c} x^2 = c^2$$ Let $\epsilon>0$ , we know that $$x^2-c^2=(x-c)(x+c)$$ $$|x^2-c^2|=|x-c||x+c|$$ If $$|x-c|<2 \implies -2<|x|-|c|<2 (by\ \ triangular \ \ property)$$ $$|c|-2<|x|<2+|c|$$ By triangular property, we have $$|x+c|\le |x|+|c|<2+|c|+|c|$$ $$|x+c|<2+2|c|$$ So, $$|x^2-c^2|<2.(2+2|c|)$$ $$|x^2-c^2|<(4+4|c|)$$ $$|x^2-c^2|<(2+2|c|)(|x-c|)$$ If $$|x-c|<\frac{\epsilon}{2+2|c|}$$ Then $$|x^2-c^2|<\epsilon$$ Define $$\delta (\epsilon) = inf \{2, \frac{\epsilon}{2+2|c|}\}$$ Then if $|x-c|< \delta \implies |x^2-c^2|<\epsilon$

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