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### Course: Differential Calculus>Unit 1

Lesson 4: Formal definition of limits (epsilon-delta)

# Formal definition of limits Part 4: using the definition

Explore the epsilon-delta definition of limits in calculus, as we rigorously prove a limit exists for a piecewise function. Dive into the process of defining delta as a function of epsilon, and learn how to apply this concept to validate limits with precision. Created by Sal Khan.

## Want to join the conversation?

• the video was really awesome, thank you so much

just a little question:

is this "ε = 2∂" true for any kind of function or just this example??
• Just this function! Because he defined f(x) to be 2x when x is not 5. Another function f(x) will yield a different function for ε.
• I'm so lost from this video
• At -, Sal says that x is not equal to 5 as if the expression |x-c| < ∂ exactly implies that x is within range of c but not equal to c. For the latter proposition, shouldn't there also be a 0 < at left-hand side of the inequality ? ( 0 < |x-c| < ∂ ). Because we are trying to approach c, not to get there, given the definition of limit.
• the inequality shown @ will include c. The inequality should read:
0<|x-c|<delta. When you remove the absolute value braces to evaluate
this expression, you will create 2 inequality expressions: one describing
the range x<c, the other the range x<c. The value x=c is excluded.
• where did the 5 come from in |x-5| ?
• In the last video, Sal wrote down the equation |x - c| < ∂, which means x is within ∂ of c. |x - 5| < ∂ is the same thing, but we are defining c to be 5.
• What would it look like if the limit you were trying to prove was actually wrong? For example, if you tried to prove that for f(x) = x^2, that as x => 3 that L = 10.
• To disprove a limit, we can show that there is some ∈>0 such that there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.

Let's say ∈<1 (because 3^2=9 and |9-10|=1).
We can always pick c=3 so that |x-c|<δ (because |x-c|=0 and 0<δ), but |f(c)-L|>∈ (because |f(c)-L|=|9-10|=1 and 1>∈).
This means that if ∈<1, there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.

Thus, we have disproved that the limit of x^2 as x approaches 3 is 10.

I hope this helps you disprove limits with the epsilon-delta definition!
• What if it's a function that grows in a non linear way, like in an exponential function?
For exemple, in the function f(x) = 2^x. Given E > 0, would the delta in the left side of a value c be different from the delta in the right side of c?
• but the proof still stands ,that's where the one sided limit proof comes from i think?
• What does he mean when he writes, '|x - c| < delta'?
• This is a formal way of writing that the difference between x and c is less than some extremely small number, delta.
• I don't understand how |2x-10|=2delta can be replaced by |fx-L|=2delta. Can someone please explain? Why are we able to get rid of the 2 on the left side of the equation?
• First off, consider the limit we're taking. It's lim (x-->5) of 2x = 10. Now, from this we see that x = x, c = 5, f(x) = 2x and L = 10. From the statement of epsilon-delta, we have |x-5|<δ (which comes from |x-c|<δ) and |2x-10|<ε (which comes from |f(x)-L|<ε. Now, take |x-5|<δ and multiply both sides by 2. We get |2x-10|<2δ. Now, 2x and 10 are basically f(x) and L. So, substituting them, we have |f(x)-L|<2δ.