341there's stuff on mathlinks you can search.
btw, do you also think terry is cute? :D
Explain Lagranges Multipliers with some examples (please ;)
atleast giv some q on lagranges multipliers (related to inequality of course)
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341
21may be this thread will remain blank
so i m adding an old doubt
prove
if f(x)≥g(x) lim x→a f(x) ≥ limx→a g(x) assume all other necessary conditions
prove using the rigorous definition
21prophet sir , à°ªà±à°²à±€à°œà± సొలà±à°µà±† థిసౠపà±à°°à°¾à°¬à±à°²à°‚ .
341We will prove that if h(x)≥0 for all x in a neighbourhood of a, then
\lim_{x\rightarrow a} h(x) =c \ge 0
Suppose to the contrary c<0.
We have that given \epsilon > 0, we can find \delta > 0 such that |x-a| <\delta \Rightarrow |h(x)-c)|<\epsilon
But setting \epsilon = \frac{c}{2}, we see that we cannot find a corresponding \delta.
Thus c≥0.
Now let h(x) = f(x)-g(x) and you have the required result
21I think we should set ε = |c|2
thanks , it became simple after defining h = f- g
now we have a proof of the squeeze theorem ,btw did u understand what i wrote in Telegu ? ;)
21For sake of completeness we have to prove that limx→a h(x) = limx→af(x) - limx→ag(x) (*)
Let limx→af(x) = m , limx→ag(x) = n
which means every ε>0 , we have some δ1>0,such that ,for all x ,if 0<|x-a|<δ1,then |f(x)-m|<ε and some δ2>0 such that if 0<|x-a|<δ2,then |g(x)-n|<ε
all the conditions remain valid if we choose δ = min{δ1,δ2} and set ε= ε/2
By modulus inequality ε = |f(x)-m| + |g(x)-n| ≥ |f(x) - g(x) - (m-n)| ,which proves (*)
341I did. it translated solve to solvay :D
My mother tongue is kannada, but the script is almost the same