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rigorously prove x^2 is continuous
- Thread starter KingAce
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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 \(\displaystyle |x^{2}-{x_{0}}^{2}|=|x+x_{0}||x-x_{0}|\)
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 \(\displaystyle |x^{2}-{x_{0}}^{2}|=|x+x_{0}||x-x_{0}|\)
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.
\(\displaystyle f(x)=x^{2}\)
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.