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

• Mar 19th 2011, 08:31 AM
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

Thanks :)
• Mar 19th 2011, 09:06 AM
emakarov
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 AM
Quote:

Originally Posted by emakarov
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.

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 PM
emakarov
You are right; this works too.
• Mar 19th 2011, 10:54 PM