How to show that $f(x)=x^2$ is continuous at $x=1$?

Question

What definition of "continuous" are you using? What have you tried? — Chris Eagle, Oct 08 '12 at 20:16
Do you know that the product of ciontinuous functions is continuous? — Hagen von Eitzen, Oct 08 '12 at 20:18

score 48 · Accepted Answer · edited Dec 11 '16 at 00:27

48

To prove the limit exists using the fundamental definition. Here is how you proceed.

We must show that for every $\epsilon >0$ there is $\delta >0$ such that if $0<|x-1|<\delta\,,$ then $|x^2-1|<\epsilon$. Finding $\delta$ is most easily accomplished by working backward. Manipulate the second inequality until it contains a term of the form $x-1$ as in the first inequality. This is easy here. First $$ |x^2-1|=|x+1||x-1| \,. $$ In the above, there is unwanted factor of $|x+1|$, that must be bounded. If we make certain that $\delta<1$ $$ |x-1|<\delta<1 \,,$$ then $$ |x-1|< \delta \implies |x-1|< 1 \implies -1<x-1<1 \,$$ Adding $2$ to the last inequality gives $$ 1<x+1<3 \implies |x+1|<3\,.$$ So, if $$ |x^2-1|=|x+1||x-1|<3|x-1|<\epsilon \implies |x-1|<\frac{\epsilon}{3}\,. $$ Now, select $\delta = \mathrm{min}\left\{ 1, \frac{\epsilon}{3}\right\} $.

Check: given $\epsilon >0$, let $\delta = \mathrm{min}\left\{ 1, \frac{\epsilon}{3}\right\} $. Then $0<|x-1|<\delta$ implies that

$$ |x^2-1|=|x+1||x-1|<3|x-1|<3 \delta\le 3 \frac{\epsilon}{3} = \epsilon.$$

edited Dec 11 '16 at 00:27

Squirtle

6,698

answered Oct 08 '12 at 21:34

Mhenni Benghorbal

47,431

1

Do you know a book, where I can find more exercises with solutions like yours? Thanks! – GniruT Dec 16 '15 at 19:12
2

It says, "select $\delta = \min{1,\frac{\epsilon}{3}}$". What if $\delta =1$? – user5826 Apr 13 '16 at 03:22
@AlJebr If $\delta = 1$ then by the definition of $\min$, we have 1<$\epsilon $/3, and thus $ 3 < \epsilon $. Thus following the last line in the post, we have $|x^2-1|<3|x-1| < 3 \delta = 3 < \epsilon $ – Evan Rosica Mar 31 '17 at 21:20
2

@AlJebr If you're curious why we need the $\delta = \min {1,\epsilon /3 }$ in the first place, let me explain the logic. The OP showed that if $|x-1| <1 $ and $ |x-1| < \epsilon/3 $ then $|x^2 -1|< \epsilon $. Note that choosing $\delta$ to be the smallest of these two bounds ${1,\epsilon /3 }$ implies the other. Ie, WLOG, if $a<b$ and $|x-x_{0}|<a$ , then $|x-x_{0}|<a<b$ , and so by the transitive property, $|x-x_{0}|<b$ holds as well. – Evan Rosica Apr 01 '17 at 10:31
@EvanRosica Then is the choice of taking $\delta$ < 1 arbitrary? Could we have chosen another condition such as $\delta$ < 2 or $\delta$ < 0.5? Also, we are able to impose the restriction that $\delta$ < 1 because the value of $\delta$ can be arbitrarily small right? – Tan Yong Boon May 21 '23 at 09:19
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@TanYongBoon Good question. My analysis is a bit rusty these days, but it looks like this argument will work for any $\delta = \min {\alpha, \frac{\epsilon}{\alpha+2}}$. E.g. Letting $\delta = \min{2,\frac{\epsilon}{4}}$ should work. – Evan Rosica May 24 '23 at 12:51
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I see, thanks! @EvanRosica – Tan Yong Boon May 25 '23 at 10:07

score 17 · Answer 2 · answered Oct 08 '12 at 20:18

If you want to know if a function is continuous, then the definition of what it takes for a function to be continuous is important. From Calculus by Varberg, Purcell, and Rigdon:

Let $f$ be defined on an open interval containing $c$. We say that $f$ is continuous at $c$ if $$\lim_{x \to c} f(x) = f(c).$$

Notice, this actually contains three parts,

$f(c)$ is defined
$\lim\limits_{x \to c} f(x)$ exists
The two values in parts 1 and 2 are equal.

So, you need to show the 3 parts of this are true with the function $f(x) = x^2$ and when $c = 1$, or figure out which part is not true.

Is $f(1)$ defined? What is it? Does $\lim\limits_{x \to 1} x^2$ exist? What is its value? Are the two values the same?

score 7 · Answer 3 · answered Jan 06 '14 at 23:43

The product of continuous functions is continuous. The function $x$ is continuous, hence also $x^2=x\cdot x$ is continuous.

The proof that the product of continuos functions is continuous, simply relies on the theorem that states the limit of the product is the product of the limits.

score 6 · Answer 4 · 2012-10-08T20:32:53.410

Let $\epsilon > 0$ be arbitrary. Choose $\delta = \sqrt{\epsilon+1}-1 > 0$. Assume that $|x-1|<\delta$. Now $|f(x)-f(x_0)|=|x^2-1|=|(x-1)(x+1)|\leq |x-1||x+1|<(\sqrt{\epsilon+1}-1)(\sqrt{\epsilon+1}-1+2)=\epsilon$

, because if $|x-1|<\delta \Leftrightarrow -\delta < x-1 < \delta|+2 \Leftrightarrow -\delta+2 < x-1+2 < \delta+2 \Leftrightarrow |x+1| < \delta+2 =\sqrt{\epsilon+1}-1+2$

then $|x+1|<\sqrt{\epsilon+1}-1+2$.

Is this right?

score 3 · Answer 5 · answered May 27 '13 at 13:18

In his 1821 text "Cours d'Analyse", Cauchy defined continuity of $y=f(x)$ by requiring that an infinitesimal $x$-increment should necessarily produce an infinitesimal change in $y$. According to this definition, if $f(x)=x^2$, then for $\alpha$ infinitesimal, the change in $y$ is precisely $f(x+\alpha)-f(x)=(x+\alpha)^2-x^2=(x+\alpha+x)(x+\alpha-x)=\alpha(2x+\alpha)$. Since $2x+\alpha$ is finite, the product $\alpha(2x+\alpha)$ is infinitesimal. Therefore $f(x)=x^2$ is continuous by definition.

How to show that $f(x)=x^2$ is continuous at $x=1$?

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