Math Is Fun Forum

mrpace

Member

Registered: 2012-08-16

Posts: 88

Challenging proof.

Prove that if lim   f(x) =   lim   g(x) = b, and if there exists a number δ > 0 such that...
                 x→a           x→a       

0 < |x − a| < δ ⇒ f(x) ≤ h(x) ≤ g(x)

then   lim  h(x) = b.
        x→a 

Worked on this for a while without getting anywhere. Any ideas???

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Challenging proof.

This is known as the sandwich theorem or squeeze theorem.

Last edited by Olinguito (2014-08-13 22:33:46)

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Bassaricyon neblina

mrpace

Member

Registered: 2012-08-16

Posts: 88

Re: Challenging proof.

Hey that's really nice. Just one thing. How would you say in English min{d1,d2,d}?

and i see where the d1 and d2 come from in the brakets above, but where does the d come from??

d=delta obv.

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Challenging proof.

is read as “minimum of delta sub one, delta sub two, and delta”. The

is from the question itself.

To clarify my working above: we take

because we need

to be less than all of

,

, and

.

We need

for

.

We need

for

.

We need

for

.

Last edited by Olinguito (2014-08-14 10:33:35)

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Bassaricyon neblina