Hello,I need help in proving :

(Note a is totally different than A )

If Limit F(x) x==>a = L (equals L ) and A<L then for an 'x' which is very close to a we get that F(x)>A

Any ideads?

Thanks :)

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- Mar 19th 2011, 08:31 AMkadmanyLimits,Prove that F(x) > A when x==> a
Hello,I need help in proving :

(Note a is totally different than A )

If Limit F(x) x==>a = L (equals L ) and A<L then for an 'x' which is very close to a we get that F(x)>A

Any ideads?

Thanks :) - Mar 19th 2011, 09:06 AMemakarov
The definition of $\displaystyle \lim_{x\to a}F(x)=L$ starts with "For every $\displaystyle \varepsilon>0$, ...". Substitute, say, $\displaystyle \varepsilon=(L-A)/2$ (or any concrete $\displaystyle \varepsilon$ that is less than $\displaystyle L-A$) and see what the rest of the definition says.

- Mar 19th 2011, 10:16 AMkadmany
Aha I see,but why its not correct to choose L - A? ;

0<|x-a|<dlta ====> |F(x)- L|< e

by |y|<m >> -m<y<m

We get ; -e<F(x)-L<e

Now e=L -A

so ==> F(x) - L > -e = -(L -A)=A - L

/ + (-L)

===> F(x)>A- L + L = A ==> F(x)> A!

Not correct ?(it is given that L -A >0,so the value of e is positive! and it fits with the requirements! )

Thank you - Mar 19th 2011, 01:00 PMemakarov
You are right; this works too.

- Mar 19th 2011, 10:54 PMkadmany