[MUSIC] Here's a claim. The limit of 2x as x approaches 10, is
20. Now, to justify this claim using an
epsilon delta argument I have to be able to prove that delta for every epsilon,
alright? You're going to demand that 2x be within epsilon of 20.
I'm going to satisfy your demand by saying x is within some number delta of
10. [SOUND] So proof, let epsilon be bigger
than zero. I don't know how small a number epsilon
you're going to choose. I'm going to set delta equal epsilon over
two. I'm going to check now that this value of
delta will work to satisfy your demand involving epsilon.
So, if zero is less than x - 10 is less than delta, right? If x is within delta
of 10, then if you multiply by 2, we see that 2 * x-10 is less than 2 delta.
But what's 2 delta? Delta is epsilon over 2.
2 delta is just epsilon. So then, 2 times the absolute value of
x-10 is less than 2 delta, which is epsilon. [SOUND] And so, 2 times the
absolute value of x-10, that's the absolute value of 2x - 20, and that's
less than epsilon. So look at what we've shown,
alright? Some epsilon bigger than zero, a corresponding value of delta is going to
be epsilon over 2. And then, I verify that if x is within 10, is within delta
of 10, then 2x is within epsilon of 20. And that's exactly what it means to say
that the limit of 2x is 20 as x approaches 10.
For every epsilon, there's some delta so that whenever x is within delta of 10,
then the function 2x is within epsilon of 20.
You can imagine trying to do this with other functions, right?
But I'll let you try that. You can try that in the exercise included
below. [MUSIC]