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

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What definition of "continuous" are you using? What have you tried? –  Chris Eagle Oct 8 '12 at 20:16
Do you know that the product of ciontinuous functions is continuous? –  Hagen von Eitzen Oct 8 '12 at 20:18
I don't understand why $6$ people have this question tagged as $favourite$? –  Awesome Mar 18 '14 at 8:28

5 Answers 5

up vote 26 down vote accepted

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

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

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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,

  1. $f(c)$ is defined
  2. $\lim\limits_{x \to c} f(x)$ exists
  3. 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?

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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?

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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.

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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.

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