Inequality

We 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

I 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 ? ;)

For sake of completeness we have to prove that lim_x→a h(x) = lim_x→af(x) - lim_x→ag(x) (*)

Let lim_x→af(x) = m , lim_x→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 (*)

I did. it translated solve to solvay :D

My mother tongue is kannada, but the script is almost the same