[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]