rigorously prove x^2 is continuous

[KingAce]

"Prove $\displaystyle x^{2}$is continuous using $\displaystyle \epsilon-{\delta}"$

Why did you delete your post?. Others may find it helpful.

 

[galactus]

Re: delta epsilon proof that f(x)=x^2 is continuous

I assume you already know the formal definition of continuity then?.

Remember the triangle inequality.

Think about \(\displaystyle |f(x)-f(x_{0})}|\).

we have $\displaystyle |x^{2}-{x_{0}}^{2}|=|x+x_{0}||x-x_{0}|$

 

[KingAce]

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[galactus]

Re: delta epsilon proof that f(x)=x^2 is continuous

$\displaystyle f(x)=x^{2}$

Let $\displaystyle \epsilon > 0$. Then a $\displaystyle {\delta} > 0$ must be found such that $\displaystyle |x-x_{0}| < {\delta}$ implies that $\displaystyle |f(x)-f(x_{0})| < {\epsilon}$ for $\displaystyle x, x_{0}\in R$.

Now, $\displaystyle |f(x)-f(x_{0})|=|x^{2}-{x_{0}}^{2}|=|(x+x_{0}||x-x_{0}|=|x+x_{0}||x-x_{0}|$

Let I be an interval of length $\displaystyle 2{\delta}, \;\ 0<{\delta}<1$ on the real line centered at $\displaystyle x_{0}$. Then, $\displaystyle |x-x_{0}|<{\delta} \;\ and \;\ |x+x_{0}|=|x-x_{0}+2x_{0}|\leq |x-x_{0}|+2|x_{0}|<1+2|x_{0}|$.

Note that $\displaystyle {\delta}$ is chosen less than 1.

Therefore, thus and hence:

$\displaystyle |x^{2}-{x_{0}}^{2}|=|x+x_{0}||x-x_{0}|\leq (1+2|x_{0}|){\delta}$.

Thus, delta is to be chosen so that $\displaystyle (1+2|x_{0}|){\delta}<{\epsilon} \;\ or \;\ {\delta}<\frac{\epsilon}{1+2|x_{0}|}$

Since delta can be found both for all epsilon > 0 and for all $\displaystyle x_{0}\in R$, it follows that f is continuous everywhere.