Limits,Prove that F(x) > A when x==> a

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March 19th 2011, 08:31 AM
kadmany

Limits,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 :)
March 19th 2011, 09:06 AM
emakarov

The definition of $\lim_{x\to a}F(x)=L$ starts with "For every $\varepsilon>0$ , ...". Substitute, say, $\varepsilon=(L-A)/2$ (or any concrete $\varepsilon$ that is less than $L-A$ ) and see what the rest of the definition says.
March 19th 2011, 10:16 AM
kadmany

Quote:

Originally Posted by emakarov

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

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
March 19th 2011, 01:00 PM
emakarov

You are right; this works too.
March 19th 2011, 10:54 PM
kadmany

Quote:

Originally Posted by emakarov

You are right; this works too.

Thank you issue been solved!