How to show that $f(x)=x^2$ is continuous at $x=1$? How to show that $f(x)=x^2$ is continuous at $x=1$?
 A: 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. 
A: 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?
A: 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
$$ |x^2-1|=|x+1||x-1|<3|x-1|<3 \delta\le 3 \frac{\epsilon}{3} = \epsilon.$$  
A: 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.
A: 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?
