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[MUSIC] Here's a claim. 
The limit of 2x as x approaches 10, is 

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20. 
Now, to justify this claim using an 

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epsilon delta argument I have to be able 
to prove that delta for every epsilon, 

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alright? You're going to demand that 2x 
be within epsilon of 20. 

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I'm going to satisfy your demand by 
saying x is within some number delta of 

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10. 
[SOUND] So proof, let epsilon be bigger 

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than zero. 
I don't know how small a number epsilon 

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you're going to choose. 
I'm going to set delta equal epsilon over 

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two. 
I'm going to check now that this value of 

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delta will work to satisfy your demand 
involving epsilon. 

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So, if zero is less than x - 10 is less 
than delta, right? If x is within delta 

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of 10, then if you multiply by 2, we see 
that 2 * x-10 is less than 2 delta. 

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But what's 2 delta? Delta is epsilon over 
2. 

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2 delta is just epsilon. 
So then, 2 times the absolute value of 

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x-10 is less than 2 delta, which is 
epsilon. [SOUND] And so, 2 times the 

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absolute value of x-10, that's the 
absolute value of 2x - 20, and that's 

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less than epsilon. 
So look at what we've shown, 

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alright? Some epsilon bigger than zero, a 
corresponding value of delta is going to 

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be epsilon over 2. And then, I verify 
that if x is within 10, is within delta 

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of 10, then 2x is within epsilon of 20. 
And that's exactly what it means to say 

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that the limit of 2x is 20 as x 
approaches 10. 

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For every epsilon, there's some delta so 
that whenever x is within delta of 10, 

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then the function 2x is within epsilon of 
20. 

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You can imagine trying to do this with 
other functions, right? 

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But I'll let you try that. 
You can try that in the exercise included 

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below. 
[MUSIC]